Newton’s first law of motion – problems and solutions

1. A person is in an elevator that moving upward at a constant velocity. The weight of the person is 800 N. Immediately the elevator rope is broke, so the elevator falls. Determine the normal force acted by elevator’s floor to the person just before and after the elevator’s rope broke.

A. 800 N and 0

B. 800 N and 800 N

C. 1600 N and 0

D. 1600 N and 800 N

Known :

Weight (w) = 800 Newton

Wanted: The normal force (N)

Solution :

Before the elevator’s rope broke

When the person stands on the floor of the elevator, weight acts on the person where the direction of the person is downward. That person at rest so that there is must a normal force acts on the person, where the direction of the normal force is upward and the magnitude of the normal force same as the magnitude of the weight.

Newton's first law of motion – problems and solutions 1Because the person is at rest in the elevator and the elevator moves at a constant speed (no acceleration), so there is no net force act on the person.

F = 0

N – w = 0

N = w

N = 800 Newton

After the elevator’s rope broke

After the elevator’s rope broke, elevator and the person free fall together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.

The correct answer is A.

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2. A block with a mass of 20 gram moves at a constant velocity on a rough horizontal floor at a constant velocity if there is an external force of 2 N acts on the block. Determine the magnitude of the friction force experienced by the block.

A. 0.3 N

B. 1.4 N

C. 2.0 N

D. 3.6 N

Known :

Mass (m) = 20 gram

Force (F) = 2 Newton

Wanted: Magnitude of friction force experienced by the block.

Solution :

Newton's first law of motion – problems and solutions 2Based on Newton’s first law of motion, if a block moves at a constant velocity, then the block has no acceleration. The block moves at a constant velocity, and there is no acceleration if :

– The magnitude of friction force (Ffric) same as the magnitude of the external force (F)

– The friction force (Ffric) has opposite direction with the external force (F)

Apply Newton’s first law of motion :

F = 0

F – Ffric = 0

F = Ffric

Ffric = 2 Newton

The correct answer is C.

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3. A smooth inclined plane with the length of 0.6 m and height of 0.4 m. A block with the weight of 1350 N will move upward using the inclined plane. Determine the magnitude of force need to move the block.

A. 100 N

B. 300 N

C. 600 N

D. 900 N

Known :

Weight of block (w) = 1350 Newton

hyp = 0.6 m

opp = 0.4 m

Wanted : The minimum force

Solution :

Newton's first law of motion – problems and solutions 4hyp = ac = 0.6 m

opp = bc = 0.4 m

Sin θ = bc / ac = 0.4 / 0.6 = 4/6 = 2/3

Based on Newton;s first law of motion, the block start to moves upward then the external force (F) minimal same as the horizontal component of weight (wx).

F = 0

F – wx = 0

F = wx

If F = wx then object start to moving upward at constant velocity.

wx = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton

The correct answer is D.

4. Three forces, F1 = 22 N, F2 = 18 N and F3 = 40 N act on a block. Which figure describes Newton’s first law.

Newton's first law and Newton's second law 1

Solution :

Newton’s first law : Net force (ΣF) = 0.

A. F1 + F2 – F3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0

B. F2 + F3 – F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

C. F2 + F3 – F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

D. F1 + F3 – F2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)

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