# Newton’s laws of motion – problems and solutions

1. If no net force acts on an object, then :

(1) the object is not accelerated

(2) object at rest

(3) the change of velocity of an object = 0

(4) the object can not travels at a constant velocity

Which statement is correct.

Solution

The correct statement :

(1) The object is not accelerated

The net force causes acceleration of an object. So if no net force then objects is not accelerated.

(2) Object at rest

Newton’s first law of motion states that if no net force acts on an object then an object always at rest or the object is always traveling at a constant velocity.

(3) The change of velocity of an object = 0

Change of velocity = acceleration. No change of velocity means no acceleration. If no acceleration then no net force acts on an object.

2. The weight of a person in an elevator at rest = 500 N. Acceleration due to gravity is 10 m/s^{2}. When lift accelerated, the tension force is 750 N. What is the acceleration of lift.

__Known :__

Person’s weight (w) = 500 Newton = 500 kg m s^{–2 }(lift at rest)

Acceleration of gravity (g) = 10 m s^{–2}

Person’s mass (m) = 500 / 10 = 50 kg

Tension force (T) = 750 N (lift accelerated)

Elevator’s mass ignored.

__Wanted:__ Acceleration of elevator

__Solution :__

Elevator at rest, no acceleration (a = 0). Force acts upward has plus sign and force acts downward has minus sign.

ΣF = m a

T – w = 0

T = w

T = 500 Newton

If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.

The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.

T – w = m a

750 – 500 = 50 a

250 = 50 a

a = 250 / 50

a = 5.0 m s^{–2}

3. An 60-kg person in an elevator accelerated downward at 3 m/s^{2}. If acceleration due to gravity is 10 m/s^{2}, what is the normal force exerted by elevator’s floor on person.

__Known :__

Mass (m) = 60 kg

Acceleration of person and elevator (a) = 3 m/s^{2}

Acceleration due to gravity (g) = 10 m/s^{2}

Weight (w) = m g = (60)(10) = 600 Newton

__Wanted:__ The normal force (N)

__Solution :__

There are two forces acts on the person in the elevator, that is weight (w) of person and the normal force (N) exerted by the floor on the person. There are three vector quantities, that is weight (w), normal force (N) and acceleration of elevator, where weight acts downward, the normal force acts upward, acceleration of elevator is downward. Vector quantities that act downward have plus sign and vector quantities that act upward have minus sign.

∑F = m a

w – N = (60)(3)

600 – N = 180

N = 600 – 180

N = 420 Newton

4. A 40-kg object in an elevator accelerated upward. If the elevator’s floor exerts 520 N on object and acceleration due to gravity is 10 m/s^{2}

__Known :__

Mass (m) = 40 kg

Normal force (N) = 520 N

Acceleration due to gravity (g) = 10 m/s^{2}

weight (w) = m g = (40)(10) = 400 N

__Wanted :__ Acceleration of elevator

__Solution :__

∑F = m a

400 – 520 = (40)(a)

-120 = (40)(a)

a = -120/40

a = -3 m/s^{2}

Acceleration of elevator is 3 m/s^{2}. Minus sign indicates that elevator travels upward.

5. An 60-kg object in an elevator accelerated downward at 3 m/s^{2}. What is the force exerted by object on the elevator’s floor.

__Known :__

Mass (m) = 60 kg

Weight (w) = m g = (60 kg)(10 m/s^{2}) = 600 kg m/s^{2 }= 600 Newton

Acceleration of elevator (a) = 3 m/s^{2}, downward

__Wanted :__ Force exerted by object on the elevator’s floor.

__Solution :__

*Elevator accelerated downward at 3 m/s**. *Force acts downward has plus sign and force acts upward has minus sign.

w – N = m a

N = w – m a

N = 600 – (60)(3)

N = 600 – 180

N = 420 Newton

Force exerted by the object on the elevator’s floor = 420 N.

6. Two blocks 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 block A is 6 kg and mass of block B is 2 kg. Acceleration due to gravity is 10 m/s^{2}. What is the tension force?

__Known :__

m_{A} = 6 kg, m_{B} = 2 kg, g = 10 m/s^{2}

w_{A} = m_{A} g = (6 kg)(10 m/s^{2}) = 60 kg m/s^{2}

w_{B} = m_{B} g = (2 kg)(10 m/s^{2}) = 20 kg m/s^{2}

__Wanted :__ tension force (T) ?

__Solution :__

w_{A} > w_{B} so that m_{A} moves downward, m_{B} moves upward.

Newton’s second law :

ΣF = m a

w_{A} – w_{B} = (m_{A} + m_{B}) a

60 – 20 = (6 + 2) a

40 = (8) a

a = 40 / 8 = 5 m/s^{2}

Tension force :

m_{A} moves downward :

w_{A} – T_{A} = m_{A} a

60 – T_{A} = (6)(5)

60 – T_{A} = 30

T_{A} = 60 – 30

T_{2} = 30 Newton

m_{B} moves upward :

T_{B} – w_{B} = m_{B} a

T_{B} – 20 = (2)(5)

T_{B} – 20 = 10

T_{B} = 10 + 20

T_{1} = 30 Newton

Tension force (T) = 30 Newton.

7. Mass of object A = 5 kg, acceleration due to gravity (g) = 10 m s^{-2}. Object A moves downward at 2.5 m.s^{-2}. What is the mass of B ?

__Known :__

Mass A (m_{A}) = 5 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Acceleration of object A (a) = 2.5 m/s^{2}

Weight A (w_{A}) = (m_{A})(g) = (5)(10) = 50 Newton

__Wanted :__ Mass of object B (m_{B})

__Solution :__

Block A moves downward so weight of object A (w_{A}) larger than weight of object B (w_{B}).

Apply Newton’s second law :

ΣF = m a

w_{A }– w_{B} = (m_{A} + m_{B}) a

50 – (m_{B})(10) = (5 + m_{B}) (2.5)

50 – 10 m_{B} = 12.5 + 2.5 m_{B}

50 – 12.5 = 2.5 m_{B} + 10 m_{B}

37.5 = 12.5 m_{B}

m_{B} = 3 kg

8. Acceleration due to gravity is 10 m/s^{2}. What is the tension force.

__Known :__

Mass of object 1 (m_{1}) = 2 kg

Mass of object 2 (m_{2}) = 3 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Weight 1 (w_{1}) = (m_{1})(g) = (2 kg)(10 m/s^{2}) = 20 kg m/s^{2}

Weight 2 (w_{2}) = (m_{2})(g) = (3 kg)(10 m/s^{2}) = 30 kg m/s^{2}

__Wanted : __ tension force (T)

__Solution :__

w_{2} > w_{1} so that m_{2} moves downward and m_{1} moves upward.

Newton’s second law of motion :

ΣF = m a

w_{2} – w_{1} = (m_{1} + m_{2}) a

30 – 20 = (2 + 3 ) a

10 = (5) a

a = 10 / 5 = 2 m/s^{2}.

Tension force ?

m_{2} moves downward

w_{2} – T_{2} = m_{2} a

30 – T_{2} = (3)(2)

30 – T_{2} = 6

T_{2} = 30 – 6

T_{2} = 24 Newton

m_{1} moves upward

T_{1} – w_{1} = m_{1} a

T_{1} – 20 = (2)(2)

T_{1} – 20 = 4

T_{1} = 20 + 4

T_{1} = 24 Newton

Tension force (T) = 24 Newton.