Solved Problems in Basic Physics

Uniform motion in a horizontal circle – problems and solutions

1. A 0.2-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1 meter and ball’s maximum speed is 10 rpm. What is the magnitude of the centripetal acceleration and the magnitude of the tension force?

Known :

Mass (m) = 0.2 kg

Radius (r) = 1 m

Angular velocity (ω) = 10 rev/min = 10 rev/60 s = 0.17 rev/s = (0.17)(6.28 rad)/s = 1 rad/s

Velocity (v) = r ω = (1 m)(1 rad/s) = 1 m/s

Wanted : a_s dan ΣF

Solution :

(a) The magnitude of the centripetal acceleration

(b) The magnitude of the tension force

ΣF = m a

T = m a_s

T = (0.2 kg)(1 m/s²)

T = 0.2 kg m/s²

T = 0.2 N

2. A 1-kg ball at the end of a string is revolving uniformly in a horizontal circle of radius 1 m. The cord will break when the tension in it exceeds 100 N. What is the maximum speed the ball can have?

Known : Uniform motion in a horizontal circle – problems and solutions 2

Mass (m) = 1 kg

Radius (r) = 1 meter

Tension force (T) = centripetal force (ΣF) = 100 N

Wanted: v maximum

Solution :

Uniform motion in a horizontal circle – problems and solutions 3

[wpdm_package id=’499′]

Rounding a banked curve – dynamics of circular motion problems and solutions

1. A car rounding a banked curve. What is an angle for the road which has a curve of radius 60 meters with a design speed of 20 m/ s? Assume there is no friction between car and road.

Solution

Rounding a banked curve – dynamics of cicular motion problems and solutions 1 N = normal force

N sin θ = horizontal component of the normal force

N cos θ = vertical component of the normal force

w = m g = the weight of the car

The road is designed to be banked to eliminate dependence on friction.

The net horizontal force, the horizontal component of the normal force (N sin θ), required to keep the car moving in a circle around the curve.

We choose x-axis as horizontal and y-axis as vertical, so that centripetal acceleration, a_R, is along the horizontal direction. In the horizontal direction, the only force is the horizontal component of the normal force (N sin θ), needed to produce the centripetal acceleration. N sin θ = centripetal force.

Apply Newton’s law of motion in the vertical direction :

Rounding a banked curve – dynamics of cicular motion problems and solutions 5

Apply Newton’s law of motion in the horizontal direction :

Rounding a banked curve – dynamics of cicular motion problems and solutions 7

Substituting N in equation 1 into N in equation 2 :

Rounding a banked curve – dynamics of cicular motion problems and solutions 1

[wpdm_package id=’497′]

Rounding a flat curve – dynamics of circular motion problems and solutions

1. A 2000-kg car rounds a curve on a flat road of radius 150 m. The coefficient of static friction is 0.5. Determine the maximum speed so the car follows the curve and not skid. Acceleration due to gravity = 10 m/s².

Known :

Mass (m) = 2000 kg

Radius (r) = 150 meters

Coefficient of static friction (μ_s) = 0.5

Weight (w) = m g = (2000 kg)(10 m/s²) = 20,000 kg m/s²= 20,000 N

Force of static friction (F_s) = μ_sN = μ_sw = (0.7)(20,000 N) = 14,000 N

Wanted : v

Solution :

Rounding a flat curve – dynamics of cicular motion problems and solutions 1

[wpdm_package id=’496′]

Two bodies with the same magnitude of acceleration – Application of Newton’s law of motion problems and solutions

1. Two masses m₁= 2 kg and m₂ = 5 kg are on inclined plane and are connected together by a string as shown in the figure. The coefficient of the kinetic friction between m₁ and incline is 0.2 and the coefficient of the kinetic friction between m₂ and incline is 0.1.

(a) Determine their acceleration

(b) Determine the tension force

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 1

Known :

Mass 1 (m₁) = 2 kg

Mass 2 (m₂) = 4 kg

Coefficient of the kinetic friction between m₁ and inclined plane (μ_k1) = 0.2

Coefficient of the kinetic friction between m₂ and inclined plane (μ_k2) = 0.1

Acceleration due to gravity (g) = 9.8 m/s²

a) The magnitude and direction of the acceleration

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 2

w₁ = weight 1 = m₁g = (2 kg)(9.8 m/s²) = 19.6 Newton

w_1x = w₁ sin 30^o = (19.6 N)(0.5) = 9.8 Newton

w_1y = w₁ cos 30^o= (19.6 N)(0.87) = 17 Newton

N₁= The normal force on m₁ = w_1y = 17 Newton

F_k1 = The force of the kinetic friction on m₁ = μ_k1N₁ = (0.2)(17 N) = 3.4 Newton

———

w₂ = weight 2 = m₂g = (4 kg)(9.8 m/s²) = 39.2 Newton

w_2x = w₂sin 60^o = (39.2 N)(0.87) = 34.1 Newton

w_2y = w₂ cos 60^o= (39.2 N)(0.5) = 19.6 Newton

N₂ = The normal force on m₂ = w_2y= 19.6 Newton

F_k2= The force of the kinetic friction on m₂ = μ_k2 N₂ = (0.1)(19.6 N) = 1.96 Newton

———

The magnitude of the acceleration :

∑F_x = m a_x

w_2x> w_1x so direction of the acceleration is the same as direction of w_2x.

Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.

w_2x– F_k2 – T₂ + T₁ – w_1x – F_k1= (m₁ + m₂) a_x

w_2x – F_k2 – w_1x – F_k1 = (m₁ + m₂ ) a_x

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a_x

18.94 N = (6 kg) a_x

a_x = 18.94 N : 6 kg

a_x= 3.16 m/s²

Magnitude of the acceleration = 3.16 m/s² . Direction of the acceleration = direction of T₁= direction of w_2x

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w_2x– F_k2– T₂ = m₂a_x

34.1 N – 1.96 N – T₂= (4 kg)(3.16 m/s²)

32.14 N – T₂ = 12.64 N

T₂= 32.14 N – 12.64 N = 19.5 Newton

The tension force = T = T₁ = T₂ = 19.5 Newton

2. m₁= 4 kg, m₂ = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m₁ and m₂ (c) magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 3

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 4

w₁ = m₁g = (4 kg)(9.8 m/s²) = 39.2 Newton

w₂= m₂ g = (2 kg)(9.8 m/s²) = 19.6 Newton

a) Magnitude and direction of the acceleration

∑F_y = m a_y

w₁> w₂so the direction of the object is same as the direction of the weight 1 (w₁). Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.

w₁– T₁ + T₂ – w₂= (m₁ + m₂) a_y

w₁ – w₂ = (m₁ + m₂) a_y

39.2 N – 19.6 N = (4 kg + 2 kg) a_y

19.6 N = (6 kg) a_y

a_y= 19.6 N : 6 kg

a_y = 3.26 m/s²

Magnitude of acceleration = 3.26 m/s². Direction of acceleration = direction of w₁ .

b) Magnitude of tension force which connecting m₁ and m₂

Apply Newton’s second law on m₂ :

∑F_y = m a_y

w₁ – T₁ = m₁a_y

39.2 N – T₁= (4 kg)( 3.26 m/s²)

39.2 N – T₁ = 13.04 N

T₁= 39.2 N – 13.04 N

T₁ = 26.16 Newton

Magnitude of the tension force which connection objects = T = T₁ = T₂ = 26.16 Newton

c) Magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 5 Pulley is at rest :

∑F_y= m a_y —— a_y= 0

∑F_y= 0

Upward force are positive, downward forces are negative :

T₃ – T₁ – T₂ = 0

T₃= T₁ + T₂

T₁ and T₂ have the same magnitude, T₁ = T₂ = T = 26.16 N :

T₃ = 2T = 2(26.16 N) = 52.32 Newton

3. Block 1 (m₁ = 10 kg) and block 2 (m₂= 15 kg) connected by a cord over frictionless pulley. Coefficient of the static friction between the block 2 with incline = 0.6. The coefficient of the kinetic friction between the block 2 with incline = 0.42. Determine (a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward (b) Determine the magnitude of the tension force.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 6

Solution

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 7

w₁ = The weight of the block 1 = m₁ g = (10 kg)(9.8 m/s²) = 98 Newton

w₂ = The weight of the block 2 = m₂ g = (15 kg)(9.8 m/s²) = 147 Newton

w_2y = w₂ cos 30^o = (147 N)(0.87) = 127.89 Newton

w_2x = w₂sin 30^o = (147 N)(0.5) = 73.5 Newton

N₂ = The normal force on the block 2 = w_2y= 127.89 Newton

F_k2 = The force of the kinetic friction on the block 2 = μ_k2 N₂ = (0.42)(127.89 N) = 53.7 Newton

F_s2 = The force of the static friction on the block 2 = μ_s2 N₂ = (0.6)(127.89 N) = 76.7 Newton

a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward

∑F_x = m a_x —— a_x = 0

∑F_x = 0

Upward forces and rightward forces are positive, downward forces and leftward forces are negative.

F – F_k2 – w_2x – w₁ – T₂ + T₁ = 0

F – F_k2 – w_2x– w₁ = 0

F = F_k2+ w_2x + w₁

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Newton

b) The magnitude of the tension force

Apply Newton’s law of the motion on the block 1 :

∑F_y = m a_y—— a_y= 0

∑F_y = 0

T₁ – w₁ = 0

T₁ = w₁ = 98 Newton

Apply Newton’s law of the motion on the block 2 :

F – F_k2 – w_2x – T₂ = 0

T₂= F – F_k2– w_2x

T₂= 225.2 N – 53.7 N – 73.5 N

T₂ = 98 Newton

Magnitude of the tension force = T₁ = T₂ = T = 98 Newton

4. Block 1 (m₁ = 16 kg) lies on a horizontal surface and the block 2 (m₂ = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m₃ = 5 kg) lies on the block 2. The coefficient of the kinetic friction between the block 2 and the horizontal surface is 0,4. The coefficient of the static friction between the block 2 with the block 3 is 0,3.

(a) When the system is released from rest, the block 3 and the block 2 still slide together ?

(b) If there is no block 3, what is the acceleration of the block 1 and the block 2 ?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 8

Solution :

a) When the system is released from rest, the block 3 and the block 2 still slide together?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 9

w₁ = The weight of the block 1 = m₁ g = (16 kg)(9.8 m/s²) = 156.8 Newton

w_1x = w₁ sin 60^o = (156.8 N)(0.87) = 136.4 Newton

w_1y = w₁ cos 60^o = (156.8 N)(0.5) = 78.4 Newton

N₁ = The normal force exerted on the block 1 by the inclined plane = w_1y = 78.4 Newton

w₃ = The weight of the block 3 = m₃ g = (5 kg)(9.8 m/s²) = 49 Newton

N₂₃ = The normal force exerted on the block 3 bythe block 2 = w₃= 49 Newton

N₃₂ = The normal force exerted on the block 2 by the block 3 = N₂₃ = w₃ = 49 Newton

(N₂₃and N₃₂ are action-reaction pair)

F_s23 = The force of the static friction exerted on the block 3 by the block 2 = μ_s N₂₃= (0.3)(49 N) = 14.7 Newton

F_s32= The force of the static friction exerted on th block 2 by the block 3 = F_s₂₃ = 14.7 Newton

(F_s23 and F_s32 are action-reaction pair)

w₂ = The weight of the block 2 = m₂ g = (12 kg)(9.8 m/s²) = 117.6 Newton

N₂ = The normal force exerted on the object 2 by the horizontal surface = w₂ + N₃₂ = 117.6 Newton + 49

Newton = 166.6 Newton

F_k2= The force of the kinetic friction on the block 2 = μ_k N₂ = (0.4)(166.6 N) = 66.64 Newton

Apply Newton’s law of motion on the block 3 :

∑F_x = m a_x

F_s23=m₃a_x

—–> F_s23 = μ_sN₂₃= μ_sw₃= μ_sm₃g

μ_s m₃g = m₃a_x

μ_s g = a_x

a_x = (0.3)(9.8 m/s²) = 2.94 m/s²

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m/s².

Now we calculate the magnitude of the system’s acceleration after released from rest.

The direction of the block displacement = the direction of the block’s acceleration = the direction of T₂= the direction of w_1x.

∑F_x = m a_x

w_1x – T₁+ T₂– F_k2 – F_s32 + F_s23 = (m₁ + m₂ + m₃) a_x

w_1x – F_k2 = (m₁ + m₂+ m₃ ) a_x

136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) a_x

69.76 N = (33 kg) a_x

a_x = 2.11 m/s²

a_x is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T₂ or direction of w_1x.

The magnitude of the acceleration is 2.11 m/s² , lower than 2.94 m/s²so we can conclude that block 3 and block 2 still slide together after released from rest.

b) The magnitude of the acceleration of the block 1 and the block 2

∑F_x= m a_x

w_1x – F_k2= (m₁ + m₂) a_x

—–> F_k2 = μ_k N₂= μ_k w₂ = μ_k m₂g = (0.4)(12 kg)(9.8 m/s²) = 47.04 Newton

136.4 N – 47.04 N = (16 kg + 12 kg) a_x

89.36 N = (28 kg) a_x

a_x= 89.36 N : 28 kg = 3.19 m/s²

[wpdm_package id=’493′]

Equilibrium of the bodies on a inclined plane – application of the Newton’s first law problems and solutions

1. A 2-kg block lies on a rough inclined plane at an angle 37^oto the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37^o = 0.6, cos 37^o = 0.8, g = 10 m.s^-2, µ_k = 0.2)

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 1 Known :

Mass (m) = 2 kg

Acceleration due to gravity (g) = 10 m/s²

Block’s weight (w) = m g = (2)(10) = 20 Newton

Sin 37^o = 0.6

Cos 37^o = 0.8

Coefficient of the kinetic friction (µ_k) = 0.2

The y-component of the weight (w_y) = w cos 37^o= (20)(0.8) = 16 Newton

The x-component of the weight (w_x) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newton

the normal force (N) = w_y= 16 Newton

Wanted : The external force (F)

Solution :

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 2 w_x = 12 Newton

The force of the kinetic friction (f_k) = µ_k N = (0.1)(16) = 1.6 Newton

The magnitude of the external force F exerted on the block :

F + f_k – w_x= 0

F = w_x– f_k

F = 12 – 1.6

F = 10.4 Newton

The external force F greater than 10.4 Newton.

2. Mass of a block = 2 kg, coefficient of static friction µ_s= 0.4 and θ = 45^o. Determine the magnitude of the force F so the block start to slides up.

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 3 Known :

The coefficient of the static friction (µ_s) = 0.4

Angle (θ) = 45^o

Acceleration due to gravity (g) = 10 m/s²

Block’s mass (m) = 2 kilogram

Block’s weight (w) = m g = (2 kg)(10 m/s²) = 20 kg m/s² = 20 Newton

The x-component of the weight (w_x) = w sin θ = (20)(sin 45) = (20)(0.5√2) = 10√2 Newton

The y-component of the weight (w_y) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 Newton

Wanted : The magnitude of the force F

Solution :

Equilibrium of bodies on inclined plane – application of Newton's first law problems and solutions 4 Block starts to slide up, if F ≥ w_x + f_s.

The x-component of the weight :

w_x = 10√2 Newton

the y-component of the weight :

w_y = 10√2 Newton

The normal force :

N = w_y = 10√2 Newton

The force of the static friction :

f_s = µ_sN = (0,4)(10√2) = 4√2

The magnitude of the force F so that the block starts to slide up :

F ≥ w_x + f_s

F ≥ 10√2 + 4√2

F ≥ 14√2 Newton

[wpdm_package id=’492′]

Equilibrium of bodies connected by cord and pulley – application of Newton’s first law problems and solutions

1. A box of mass 5 kg is on an inclined plane at an angle 30^o. The box supported by a cord. Determine the tension force (T) and the normal force (N)!

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 1

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 2 ∑F_x = 0

T – w sin 30^o = 0

T = w sin 30^o

T = (5 kg)(9.8 m/s²) sin 30^o

T = (49)(0.5)

T = 24.5 Newton

∑F_y = 0

N – w cos 30^o = 0

N = w cos 30^o

N = (49)(0.87)

N = 43 Newton

2. Two objects of mass m₁ = m₂ = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T₁ and T₂.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 3

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 4

(a) Free-body diagram for object 1 (b) Free-body diagram for object 2

Apply Newton’s first law to object 1 :

∑F_y = 0

T₁ – w₁ = 0

T₁ = w₁ = m₁g = (2 kg)(9.8 m/s²) = 19.6 N

Apply Newton’s first law to object 2 :

∑F_y = 0

T₂ – w₂ = 0

T₂ = w₂= m₂ g = (2 kg)(9.8 m/s²) = 19.6 N

T₁ = T₂= 19.6 N.

3. An object of weight w_A = 30 N and an object of weight w_B = 40 N, are attached by a lightweight cord that passes over a frictionless pulley of the negligible mass. Determine the coefficient of the maximum static friction between w_B and inclined surface, if the system is at rest.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 5

Solution

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 6

(a) Free-body diagram for object w_A (b) Free-body diagram for object w_B

Apply Newton’s first law to object w_A in vertical (y) direction :

∑F_y = 0 (no acceleration in vertical direction)

T – w_A= 0

T = w_A = 30 Newton

Apply Newton’s first law to object w_Bin vertical (y) direction :

∑F_y = 0

N – w_B cos 45^o= 0

N = w_Bcos 45^o= (40)(0.7) = 28 Newton

Apply Newton’s first law to object w_Bin horizontal (x) direction :

∑F_x= 0

F_k + w_B sin 45^o – T = 0

μ_s N + w_B sin 45^o – T = 0

μ_s (28) + (40)(0.7) – 30 = 0

μ_s (28) + 28 – 30 = 0

μ_s(28) = 30 – 28

μ_s(28) = 2

μ_s= 2 / 28

μ_s= 0.07

The coefficient of the maximum static friction between w_B and inclined surface = 0.07.

[wpdm_package id=’490′]

Particles in two-dimensional equilibrium – application of Newton’s first law problems and solutions

1. Find the tension force T₁, T₂, and T₃. Ignore cord’s mass.

Particles in two-dimensional equilibrium – application of Newton's first law problems and solutions 1

Solution

Particles in two-dimensional equilibrium – application of Newton's first law problems and solutions 2

(a) Free-body diagram for object (b) Free-body diagram for cord

Apply the Newton’s first law on the object :

ΣF_y = 0

T₁ – w = 0

T₁ = w = m g

T₁ = (5 kg)(9.8 m/s²)

T₁= 49 kg m/s²

T₁ = 49 N

Apply Newton’s first law on the cord :

∑F_x = 0

T_3x – T _2x= 0

T₃cos 30^o – T₂ cos 40^o= 0

0.87 T₃– 0.77 T₂= 0

0.87 T₃ = 0.77 T₂

T₂= 0.87 T₃/ 0.77 = 1.1 T₃———- Equation 1

——

∑F_y = 0

T_3y + T_2y – T_1y = 0

T₃sin 30^o + T₂sin 40^o – T₁ = 0

0.5 T₃+ 0.64 T₂– 49 N = 0 ———- Equation 2

Substituting T₂in the equation 2 into the equation 2 :

0.5 T₃+ 0.64 (1.1 T₃) – 49 N = 0

0.5 T₃ + 0.70 T₃– 49 = 0

1.2 T₃ – 49 = 0

1.2 T₃= 49

T₃ = 49 / 1.2

T₃ = 41 N

———

T₂ = 1.1 T₃

T₂ = (1.1)(40.8 N)

T₂ = 45 N

[wpdm_package id=’488′]

Particles in the one-dimensional equilibrium – application of the Newton’s first law problems and solutions

^{1. Mass of an object, m = 10 kg, supported by a cord. Find the tension in the cord!}g = 10 m/s²

Particles in one-dimensional equilibrium – application of Newton's first law problems and solutions 1 Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s²

Wanted : The tension force (T)

Solution :

ΣF_y = 0

T – w = 0

T = w

T = m g

T = (10 kg)(10 m/s²) = 100 kg m/s²

T = 100 Newton

2. Mass of the object is 10 kg. Find the tension in the cord….. Acceleration due to gravity = 10 m/s².

Solution

Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s².

Wanted : The tension force (T)

Solution :

^{w = weight = m g = (10 kg)(10 m/s2}) = 100 kg m/s²

T₁ = the tension force 1

_T1x = the x-component of the tension force 1 = T₁ cos 45^o = 0.7 T₁

_T1y = the y-component of the tension force 2 = T₁ sin 45^o = 0.7 T₁

T₂ = the tension force 2

_T2x = the x-component of the tension force 2 = T₂ cos 45^o = 0.7 T₂

_T2y = the y-component of the tension force 2 = T₂ sin 45^o = 0.7 T₂

The equilibrium condition ΣF = 0.

y axis :

ΣF_y = 0

T_1y + T_2y – w = 0

0.7T₁ + 0.7T₂ – 100 = 0

0.7T₁ + 0.7T₂ = 100 —– equation 1

x axis :

ΣF_x = 0

T_2x – T_1x = 0

0.7T₂ – 0.7T₁ = 0

0.7T₂ = 0.7T₁

T₂ = T₁ —– equation 2

Determine magnitude of T₁ :

0.7T₁+ 0.7T₁= 100

1.4T₁ = 100

T₁ = 100 / 1.4

T₁= 71.4 Newton

T₁ = T₂ so T₂ = 71.4 Newton

[wpdm_package id=’486′]

Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s². Find (a) The acceleration of the system (b) The tension in the cord!

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 1

Solution

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 2 Known :

Mass of the box 1 (m₁) = 2 kg

Mass of the box 2 (m₂) = 3 kg

Acceleration due to gravity (g) = 10 m/s²

Weight of the box 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Weight of the box 2 (w₂) = m₂ g = (3)(10) = 30 Newton

Solution :

(a) magnitude and direction of the acceleration

w₂> w₁ so the box 2 accelerates downward and the box 1 accelerates upward.

Forces that has the same direction with acceleration (w₂ and T₁), its sign positive. Forces that has opposite direction with acceleration (T₂ and w₁), its sign negative.

∑F = m a

w₂ – T₂ + T₁ – w₁ = (m₁ + m₂) a ——-> T₁ = T₂ = T

w₂ – T + T – w₁ = (m₁ + m₂) a

w₂ – w₁ = (m₁ + m₂) a

30 – 20 = (2 + 3) a

10 = 5 a

a = 10 / 5

a = 2 m/s²

Magnitude of the acceleration is 2 m/s².

(b) The tension force

The box 2 :

There are two forces acts on the box 2 : first, weight of the box 2 (w₂), points downward so it’s positive. Second, tension force exerted on the box 2 (T₂), points upward so it’s negative. Apply Newton’s second law of motion.

∑F = m a

w₂ – T₂ = m₂ a

30 – T₂ = (3)(2)

30 – T₂ = 6

T₂ = 30 – 6

T₂ = 24 Newton

Box 1 :

There are two forces acts on the box 1. First, weight of the box 1 (w₁), points downward so it’s negative. Second, the tension force exerted on the box 1 (T₁) points upward so it’s positive. Apply Newton’s second law of motion :

∑F = m a

T₁ – w₁ = m₁ a

T₁ – 20 = (2)(2)

T₁ – 20 = 4

T₁ = 20 + 4

T₁ = 24 Newton

Magnitude of the tension force = T₁ = T₂ = T = 24 Newton

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s², coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 3

Solution

Bodies connected by cord and pulley - application of Newton's law of motion problems and solutions 4 Known :

Mass of the object 1 (m₁) = 2 kg

Mass of the object 2 (m₂) = 4 kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of the static friction (μ_s) = 0.4

The coefficient of the kinetic friction (μ_k) = 0.3

Weight of the object 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Weight of the object 2 (w₂) = m₂ g = (4)(10) = 40 Newton

Normal force exerted on the object 1 (N) = w₁ = 20 Newton

Force of the static friction exerted on the object 1 (f_s) = μ_sN = (0.4)(20) = 8 Newton

Force of the kinetic friction exerted on the object 1 (f_k) = μ_kN = (0.3)(20) = 6 Newton

Wanted: acceleration (a)

Solution :

w₂ > f_s (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (f_k). Apply Newton’s second law of motion :

∑F = m a

w₂ – the = (m₁ + m₂) a

40 – 6 = (2 + 4) a

34 = 6 a

a = 34 / 6 = 17 / 3

a = 5.7 m/s²

Magnitude of the acceleration = 5.7 m/s²

[wpdm_package id=’484′]

Application of the Newton’s law of the motion in an elevator – problems and solutions

1. A 50-kg person in an elevator. Acceleration due to gravity = 10 m/s². Determine the normal force exerted on the object by the elevator, if :

(a) the elevator is at rest

(b) the elevator is moving downward at a constant velocity

(d) elevator accelerated downward at a constant 5 m/s²

(e) elevator in a free fall

Solution

Application of Newton's law of motion on elevator - problems and solutions 1 Known :

Person’s mass (m) = 50 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (w) = m g = (50)(10) = 500 Newton

Wanted: The normal force (N)

Solution :

(a) the elevator is at rest

The elevator is at rest so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = m a

N – w = 0

N = w

N = 500 Newton

(b) the elevator is moving downward at a constant velocity

Constant velocity so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = m a

N – w = 0

N = w

N = 500 Newton

The direction of the acceleration is upward, so we choose the positive direction as up.

N – w = m a

N = w + m a

N = 500 + (50)(5)

N = 500 + 250

N = 750 Newton

The person feels the floor pushing up harder than when the elevator is stationary or moving with a constant velocity.

If the person stands on a scale, the scale reads the magnitude of the downward force exerted by the person on the scale. By Newton’s third law, this equals the magnitude of the upward normal force exerted by the scale on the person.

(d) elevator accelerated downward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(5)

N = 500 – 250

N = 250 Newton

The person’s weight is 250 N, less than actual weight w = 500 N.

(e) elevator in a free fall

Free fall means the elevator’s acceleration is the same as the acceleration due to gravity. The magnitude of the acceleration due to gravity is 9,8 m/s², it’s direction is downward toward the center of the Earth. The speed increases linearly in time by 9,8 m/s during each second.

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(10)

N = 500 – 500

N = 0

2. Determine tension in an elevator cable. Elevator’s mass = 2000 kg.

(a) elevator is at rest

(b) elevator accelerated downward at a constant 5 m/s²

(d) elevator in a free fall

Acceleration due to gravity (g) = 10 m/s²

Solution

Application of Newton's law of motion on elevator - problems and solutions 2 Known :

Elevator’s mass (m) = 2000 kg

Acceleration of gravity (g) = 10 m/s²

weight (w) = m g = (2000)(10) = 20,000 Newton

Wanted : The tension force (T)

Solution :

(a) elevator is at rest

elevator is at rest so there is no acceleration (a = 0)

We choose the upward direction as the positive direction and the downward direction as the negative direction.

ΣF = m a

T – w = 0

T = w

T = 20,000 Newton

Tension in cable (T) = elevator’s weight (w) = 20,000 Newton

(b) elevator accelerated downward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(5)

T = 20,000 – 10,000

T = 10,000 Newton

c) elevator accelerated upward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as up.

T – w = m a

T = w + m a

T = 20,000 + (2000)(5)

T = 20,000 + 10,000

T = 30,000 Newton

(d) elevator in a free fall

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(10)

T = 20,000 – 20,000

T = 0

[wpdm_package id=’482′]