Forces & Motion — Science

Apply it in a new setting

Problem 1

A 1200 kg car speeds up from rest to 24 m/s in 8 s on a straight road. What net force acts on the car? Use a = Δv/t and F = ma.

Hints

First find the acceleration from the change in velocity and the time.

a = (24 − 0) ÷ 8.

Then apply Newton's second law: F = ma.

Answer and explanation

3600 N

Acceleration a = Δv/t = 24 ÷ 8 = 3 m/s^2.

Newton's second law: F = ma = 1200 · 3.

F = 3600 N.

Problem 2

A motorcycle traveling at 28 m/s brakes at a constant −7 m/s^2. How long does it take to come to a complete stop? Use v = v_0 + at.

Hints

At a complete stop the final velocity v = 0.

0 = 28 + (−7)t.

Solve for t: t = 28 ÷ 7.

Answer and explanation

4 s

Set v = 0 in v = v_0 + at: 0 = 28 + (−7)t.

So 7t = 28, giving t = 28 ÷ 7.

t = 4 s.

Problem 3

A stone is dropped from rest off a cliff and hits the ground 3 s later. Ignoring air resistance, how tall is the cliff? Use g = 9.8 m/s^2 and d = v_0 t + ½at^2.

Hints

Dropped from rest means the initial velocity v_0 = 0.

With v_0 = 0 the distance formula becomes d = ½at^2.

d = ½ · 9.8 · 3^2.

Answer and explanation

44.1 m

Starting from rest, v_0 = 0, so d = ½at^2.

d = ½ · 9.8 · 3^2 = 4.9 · 9.

d = 44.1 m.

Problem 4

A worker pushes a 20 kg crate across the floor with a 100 N horizontal force. Friction opposes the motion with 40 N. What is the crate's acceleration? Draw the free-body forces, find the net force, then use F = ma.

Hints

The push and friction point in opposite directions along the floor.

Net force = 100 − 40.

a = net force ÷ mass = 60 ÷ 20.

Answer and explanation

3 m/s^2

Net horizontal force = 100 − 40 = 60 N.

Newton's second law: a = F_net ÷ m = 60 ÷ 20.

a = 3 m/s^2.

Problem 5

A 1500 kg car moving at 20 m/s is brought to a stop in 5 s. Using the impulse–momentum relationship FΔt = Δ(mv), what average braking force acts on the car?

Hints

First find the change in momentum: Δp = mv_final − mv_start.

Δp = 1500 · 20 = 30000 kg·m/s (the car ends at rest).

Average force = Δp ÷ time = 30000 ÷ 5.

Answer and explanation

6000 N

Change in momentum Δp = m·Δv = 1500 · 20 = 30000 kg·m/s.

Impulse–momentum: F = Δp ÷ Δt = 30000 ÷ 5.

F = 6000 N.

Problem 6

A car speeds up uniformly from 10 m/s to 30 m/s while covering 100 m of road. What is its acceleration? Use v^2 = v_0^2 + 2ad.

Hints

This is the kinematic equation with no time in it.

30^2 = 10^2 + 2 · a · 100, so 900 = 100 + 200a.

Solve: 200a = 800.

Answer and explanation

4 m/s^2

v^2 = v_0^2 + 2ad → 30^2 = 10^2 + 2a(100).

900 = 100 + 200a, so 200a = 800.

a = 800 ÷ 200 = 4 m/s^2.

Problem 7

A person pushes horizontally on a heavy wall with 30 N of force, but the wall does not move. Which statement best explains the situation using Newton's laws?

The wall pushes back on the person with 30 N in the opposite direction; this reaction force acts on the person, not the wall.
The wall pushes back on the person with more than 30 N, which is why the wall stays still.
The wall exerts no force at all because it is not moving.
The person's push is cancelled out by the downward pull of gravity.

Hints

Newton's third law pairs act on two different objects, never on the same one.

The action and reaction forces are always equal in size and opposite in direction.

The wall stays put because it is anchored to the Earth, not because it pushes back harder.

Answer and explanation

The wall pushes back on the person with 30 N in the opposite direction; this reaction force acts on the person, not the wall.

By Newton's third law the wall pushes back on the person with an equal and opposite 30 N.

That reaction force acts on the person, while the push acts on the wall — they never cancel because they act on different objects.

The wall stays still because it is anchored to the ground, so A is correct.

Browse all courses