Solved problems in Newton’s laws of motion – Newton’s second law of motion
1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object.
Notum:
Massa (m) = 1 kg
acceleratio (a) = 5 m/s2
voluit : net force (∑F)
solution:
We use Newton’s second law to get the net force.
ΣF = ma
ΣF = (1 kg)(5 m/s2) = 5 kg m/s2 = 5 Newtoni
2. Missam of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….
Notum:
Massa (m) = 1 kg
Net force (∑F) = 2 Newton
voluit : The magnitude and direction of the acceleration (a)
solution:
a = ∑F / m
a = 2 / 1
a = 2 m/s2
The direction of the acceleration = the direction of the net force (∑F)
3. Object’s mass = 2 kg, F1 = 5 Newtoni, F2 = 3 Newton. The magnitude and direction of the acceleration is…
Notum:
Massa (m) = 2 kg
F1 = 5 Newtoni
F2 = 3 Newtoni
Quaesitum: The magnitude and direction of the acceleration (a)
solution:
net force :
ΣF = F1 - F2 = 5 – 3 = 2 Newtoni
Magnitudo accelerationis:
a = ∑F / m
a = 2 / 2
a = 1 m/s2
Direction of the acceleration = direction of the net force = direction of F1
4. Object’s mass = 2 kg, F1 = 10 Newtoni, F2 = 1 Newton. The magnitude and direction of the acceleration is…
Notum:
Massa (m) = 2 kg
F2 = 1 Newtoni
F1 = 10 Newtoni
F1x F =1 XXX coso = (10)(0.5) = 5 Newtoni
voluit : The magnitude and direction of the acceleration (a)
solution:
Vis netta:
ΣF = F1x - F2 = 5 – 1 = 4 Newtoni
Magnitudo accelerationis:
a = ∑F / m
a = 4 / 2
a = 2 m/s2
Direction of the acceleration = direction of the net force = direction of F1x
5. F1 = 10 Newtoni, F2 = 1 Newtonus, m1 = 1 kg, m2 = 2 kg. The magnitude and direction of the acceleration is…
Notum:
Massa 1 (m1) = 1 kg
Massa 2 (m2) = 2 kg
F1 = 10 Newtoni
F2 = 1 Newtoni
voluit : The magnitude and direction of the acceleration (a)
solution:
The net force :
ΣF = F1 - F2 = 10 – 1 = 9 Newtoni
Magnitudo accelerationis:
a = ∑F / (m1 m +2)
a = 9 / (1 + 2)
a = 9 / 3
a = 3 m/s2
The direction of the acceleration = the direction of the net force = direction of F1
6.
A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/s2. Determine the magnitude of friction force experienced by the block.
A. CLXXX N
B. 40 N
C. 43 N
D. 80 N
Notum:
Massa (m) = 40 kg
Vis (F) = 200 N
Acceleratio (a) = 3 m/s2
SE busca: Vis frictionis (Fg)
solution:
Aequatio Lex secunda motus Newtoni
ΣF = ma
ΣF = net force, m = mass, a = acceleration
The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).
Choose rightward as positive and leftward as negative.
ΣF = ma
F – Fg = ma
200 – Fg = (40)(3)
200 – Fg = 120
Fg = (L) - MMD
Fg = 80 Newtoni
Responsum rectum est D.
7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normalis vis exerted by block B on block A.
A. CLXXX N
B. 1.25 N
C. 2 N
D. 3 N
Notum:
Vis (F) = 5 Newtoni
Massa quadrati A (mA) = 100 grammata = 0.1 kg
Massa quadrati B (mB) = 300 grammata = 0.3 kg
Acceleratio gravitatis (g) = 10 m/s2
Pondus of block A (wA) = (0.1 kg)(10 m/s2) = 1 kg m/s2 = 1 Newtoni
Pondus quadrati B (wB) = (0.3 kg)(10 m/s2) = 3 kg m/s2 = 3 Newtoni
Quaesitum: Normal force exerted by block B to block A
solution:
There are several forces that act on both block, as shown in figure.
F = push force (act on block B)
wA = weight of block A (act on block A)
wB = weight of block B (act on block B)
NA = normal force exerted by block B on block A (Act on block A)
NA’ = normal force exerted by block A on block B (Act on block B)
Apply Newton’s second law of motion on both blocks :
ΣF = ma
F – wA - LatinB +NA - NA' = (m)A m +B) et
NA et NA’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.
F – wA - LatinB = (mA m +B) et
5 – 1 – 3 = (0.1 + 0.3) a
5 – 4 = (0.4) a
1 = (0.4) a
a = 1 / 0.4
a = 2.5 m/s2
Apply Newton’s second law of motion on block A :
ΣF = ma
NA - LatinA m =A a
NA – 1 = (0.1)(2.5)
NA - III XI =
NA V = + X
NA = 1.25 Newtoni
Responsum rectum est B.
8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.
A. 3 N upward
B. 4 N downward
C. 9 N upward
D. 9 N downward
Notum:
Weight of X (wX) = 4 Newtoni
Pull force (Fx) = 2 Newtoni
Tension force (FT) = 9 Newtoni
SE busca: Net force acts on object X
solution:
Vertically upward forces that act on object X :
The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.
Vertically downward forces that act on object X :
There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight wx and the horizontal component of force Fx.
Net force act on the object X :
FT - LatinX - Fx = 9 – 4 – 2 = 9 – 6 = 3
The net force act on the object X is 3 Newton, vertically upward.
Responsum rectum est A.
9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s2. If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.
A. 1.75 m/s2
B. 1.50 m/s2
Circa 1.00 m/s2
D. 0.88 m/s2
Notum:
Force (F) = 16 Newton = 16 kg m/s2
Acceleratio (a) = 2 m/s2
Friction force (Fpecunia) = 2 Newton = 2 kg m/s2
Quaesitum: Object’s acceleration ?
solution:
Smooth horizontal surface (no friction force) :
ΣF = ma
F = ma
16 = (m) 2
m = 16 / 2
m = 8 kg
Mass of object is 8 kilogram.
Rough horizontal surface (there is a friction force) :
ΣF = ma
F – Fpecunia = ma
16 – 2 = 8 a
14 = 8 a
a = 14 / 8
a = 1.75 m/s2
Object’s acceleration is 1.75 m/s2.
Responsum rectum est A.
10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms-2, then determine the magnitude and direction of force act by Tom.
A. 1.70 N and its direction is opposite with force acted by Andre.w
B. 1.70 N and its direction same as force acted by Andrew
C. 2.30 N and its direction is opposite with force acted by Andrew.
D. 2.30 N and its direction same as force acted by Andrew.
Notum:
Push force acted by Andrew (F1) = 5.70 Newtoni
Mass of object (m) = 2.00 kg
Acceleratio (a) = 2.00 m/s2
Quaesitum: Magnitude and direction of force acted by Tom (F2)?
solution:
Apply Newton’s second law of motion :
ΣF = ma
F1 F +2 = ma
5.70 + F2 = (2)(2)
5.70 + F2 = 4
F2 = (L) - MMD
F2 = – 1.7 Newton
Minus sign indicated that (F2) is opposite with push force act by Andrew (F1).
Responsum rectum est A.
11. If the mass of the block is the same, which figure shows the smallest acceleration?
Solutio
Net force A :
ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward
Net force B :
ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward
Net force C :
ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward
Net force D :
ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward
The equation of Newton’s second law :
ΣF = ma
a = ΣF / m
a = acceleration, ΣF = net force, m = mass
Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration.
Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.
Responsum rectum est B.
12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.
Determine the object’s acceleration.
Notum:
Mass of object (m) = 20 kg
Net force (ΣF) = 25 N + 30 N – 15 N = 40 N
SE busca: Acceleration of an object
solution:
Object’s acceleration calculated using the equation of Newton’s second law :
ΣF = ma
a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s2
13. Which statements below describes Newton’s third law?
(1) Passengers pushed forward when the bus braked suddenly
(2) Books on paper are not falling when the paper is pulled quickly
(3) When playing skateboard when the foot pushes the ground back then the skateboard will slide forward
(4) O*ars pushed backward, boats moving forward
solution:
(2) Newton’s first law
(3) Newton’s third law
(4) Newton’s third law
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- Massa et pondus
- Normalis vis
- Lex secunda motus Newtoni
- Vis frictionis
- Motus in superficie horizontali sine vi frictionis
- Motus duorum corporum eadem acceleratione in superficie horizontali aspera sub vi frictionis
- Motus in plano inclinato sine vi frictionis
- Motus in plano inclinato aspero cum vi frictionis
- Motus in ascensore
- Motus corporum funibus et trochleis connectitur
- Duo corpora eadem magnitudine accelerationis habentia
- Curva plana rotundanda – dynamica motus circularis
- Curvae inclinatae rotundatio – dynamica motus circularis
- Motus uniformis in circulo horizontali
- Vis centripeta in motu circulari uniformi
