Newtons andre bevegelseslov – problemer og løsninger

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.

Kjent:

Masse (m²) = 1 kg

Akselerasjon (a) = 5 m/s²

Ønsket : net force (∑F)

løsning:

We use Newton’s second law to get the net force.

ΣF = ma

ΣF = (1 kg)(5 m/s²) = 5 kg m/s² = 5 Newton

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Kjent:

Masse (m²) = 1 kg

Net force (∑F) = 2 Newton

Ønsket : The magnitude and direction of the acceleration (a)

løsning:

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 Newton, F₂ = 3 Newton. The magnitude and direction of the acceleration is…

Kjent:

Masse (m²) = 2 kg

F₁ = 5 Newton

F₂ = 3 Newton

Ønskes: The magnitude and direction of the acceleration (a)

løsning:

net force :

ΣF = F₁ - F₂ = 5 – 3 = 2 Newton

Størrelsen på akselerasjonen:

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 Newton, F₂ = 1 Newton. The magnitude and direction of the acceleration is…

Kjent:

Masse (m²) = 2 kg

F₂ = 1 Newton

F₁ = 10 Newton

F_1x =F₁ cos 60^o = (10)(0.5) = 5 Newton

Ønsket : The magnitude and direction of the acceleration (a)

løsning:

Netto kraft:

ΣF = F_1x - F₂ = 5 – 1 = 4 Newton

Størrelsen på akselerasjonen:

a = ∑F / m

a = 4 / 2

a = 2 m/s²

Direction of the acceleration = direction of the net force = direction of F_1x

5. F₁ = 10 Newton, F₂ = 1 Newton, m₁ = 1 kg, m²₂ = 2 kg. The magnitude and direction of the acceleration is…

Kjent:

Masse 1 (m₁) = 1 kg

Masse 2 (m₂) = 2 kg

F₁ = 10 Newton

F₂ = 1 Newton

Ønsket : The magnitude and direction of the acceleration (a)

løsning:

The net force :

ΣF = F₁ - F₂ = 10 – 1 = 9 Newton

Størrelsen på akselerasjonen:

a = ∑F / (m₁ +m₂)

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. 15 N

B. 40 N

C. 43 N

D. 80 N

Kjent:

Masse (m²) = 40 kg

Kraft (F) = 200 N

Akselerasjon (a) = 3 m/s²

Wanted: Friksjonskraft (F_g)

løsning:

Ligningen til Newtons andre bevegelseslov

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 – F_g = ma

200 – F_g = (40)(3)

200 – F_g = 120

F_g = 200 - 120

F_g = 80 Newton

Riktig svar er 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 normal kraft exerted by block B on block A.

A. 1 N

B. 1.25 N

C. 2 N

D. 3 N

Kjent:

Kraft (F) = 5 Newton

Masse av blokk A (m_A) = 100 gram = 0.1 kg

Masse av blokk B (m_B) = 300 gram = 0.3 kg

Tyngdekraftens akselerasjon (g) = 10 m/s²

Vekt of block A (w_A) = (0.1 kg)(10 m/s²) = 1 kg m/s² = 1 Newton

Vekt av blokk B (w_B) = (0.3 kg)(10 m/s²) = 3 kg m/s² = 3 Newton

Ønskes: Normal force exerted by block B to block A

løsning:

There are several forces that act on both block, as shown in figure.

F = push force (act on block B)

w_A = weight of block A (act on block A)

w_B = weight of block B (act on block B)

N_A = normal force exerted by block B on block A (Act on block A)

N_A’ = 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 – w_A - w_B + N_A - N_A’ = (m_A +m_B) En

N_A og N_A’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

F – w_A - w_B = (m_A +m_B) En

5 – 1 – 3 = (0.1 + 0.3) a

5 – 4 = (0.4) a

1 = (0.4) en

a = 1 / 0.4

a = 2.5 m/s²

Apply Newton’s second law of motion on block A :

ΣF = ma

N_A - w_A = m_A a

N_A – 1 = (0.1)(2.5)

N_A - 1 = 0.25

N_A = 1 + 0.25

N_A = 1.25 Newton

Riktig svar er 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

Kjent:

Weight of X (w_X) = 4 Newton

Pull force (F_x) = 2 Newton

Tension force (F_T) = 9 Newton

Wanted: Net force acts on object X

løsning:

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 w_x and the horizontal component of force F_x.

Net force act on the object X :

F_T - w_X - F_x = 9 – 4 – 2 = 9 – 6 = 3

The net force act on the object X is 3 Newton, vertically upward.

Det riktige svaret er 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

C. 1.00 m/s²

D. 0.88 m/s²

Kjent:

Force (F) = 16 Newton = 16 kg m/s²

Akselerasjon (a) = 2 m/s²

Friksjonskraft (F_frik) = 2 Newton = 2 kg m/s²

Ønskes: Object’s acceleration ?

løsning:

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 – F_frik = ma

16 – 2 = 8 a

14 = 8 a

a = 14 / 8

a = 1.75 m/s²

Object’s acceleration is 1.75 m/s².

Det riktige svaret er 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.

Kjent:

Push force acted by Andrew (F₁) = 5.70 Newton

Mass of object (m) = 2.00 kg

Akselerasjon (a) = 2.00 m/s²

Ønskes: Magnitude and direction of force acted by Tom (F₂)?

løsning:

Apply Newton’s second law of motion :

ΣF = ma

F₁ + F₂ = ma

5.70 + F₂ = (2)(2)

5.70 + F₂ = 4

F₂ = 4 - 5.70

F₂ = – 1.7 Newton

Minus sign indicated that (F₂) is opposite with push force act by Andrew (F₁).

Det riktige svaret er A.

11. If the mass of the block is the same, which figure shows the smallest acceleration?

Oppløsning

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.

Riktig svar er 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.

Kjent:

Mass of object (m) = 20 kg

Net force (ΣF) = 25 N + 30 N – 15 N = 40 N

Wanted: Acceleration of an object

løsning:

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) Åars pushed backward, boats moving forward

løsning:

(1) Newtons første lov

(2) Newton’s first law

(3) Newton’s third law

(4) Newton’s third law

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