Laws of Motion — Practice

Question 1

A 5 kg block is pulled by a 20 N horizontal force against 8 N of friction. Find its acceleration.

Accepted Answer

Net force = 20 - 8 = 12 N. Then a = F_net/m = 12/5 = 2.4 m/s^2, in the direction of the applied force. Note that F in the second law is always the net force, never just the applied one.

Question 2

A book rests on a table. Name the third-law reaction to the weight of the book, and say why it is not the normal force from the table.

Accepted Answer

The weight is the Earth pulling the book down, so its reaction is the book pulling the Earth up with an equal force. The normal force is not the reaction, because it acts on the same body as the weight — the book — and a third-law pair must act on two different bodies. The normal force and the weight happen to balance here, which is a separate fact.

Question 3

Why is it harder to start a heavy box moving than to keep it moving?

Accepted Answer

Because the coefficient of static friction is greater than the coefficient of kinetic friction. The maximum static friction that must be overcome to start motion, mu_s N, exceeds the kinetic friction mu_k N that opposes motion once it has begun. This is also why a crate often lurches forward the moment it breaks free.

Question 4

A 0.5 kg ball moving at 20 m/s is caught in 0.02 s. Find the average force on the hands.

Accepted Answer

Change in momentum = 0 - (0.5 x 20) = -10 kg m/s, so the magnitude of the impulse is 10 N s. Average force = |delta-p|/delta-t = 10/0.02 = 500 N. Catching over 0.20 s instead would cut this to 50 N, which is exactly why a fielder pulls the hands back.

Question 5

A 4 kg gun fires a 20 g bullet at 400 m/s. Find the recoil speed of the gun.

Accepted Answer

The net external force is zero, so total momentum is conserved. Before firing it is zero, so afterwards the two momenta must be equal and opposite: 4 x v = 0.020 x 400 = 8 kg m/s, giving v = 2 m/s backwards. The gun recoils slowly because it is 200 times heavier than the bullet.

Question 6

A block slides down a 30 degree incline with mu_k = 0.2. Find its acceleration (g = 10 m/s^2).

Accepted Answer

Resolving along the incline: a = g(sin(theta) - mu_k cos(theta)) = 10(0.5 - 0.2 x 0.866) = 10(0.5 - 0.173) = 3.27 m/s^2, directed down the slope. The mass cancels, so the answer is the same for any block on those surfaces.

Question 7

A 1200 kg car turns a curve of radius 20 m at 15 m/s. Find the friction force required.

Accepted Answer

F_c = mv^2/R = 1200 x 225/20 = 13500 N, directed toward the centre of the curve. On a level road friction is the only force available to supply it, so the road must be able to provide 13500 N or the car slides outward.

Question 8

Find the maximum speed at which a car can round an unbanked curve of radius 50 m if mu_s = 0.5 (g = 10 m/s^2).

Accepted Answer

Friction supplies the centripetal force, so mv^2/R <= mu_s mg, giving v_max = sqrt(mu_s R g) = sqrt(0.5 x 50 x 10) = sqrt(250) = 15.8 m/s. The mass cancels, so a fully loaded car and an empty one have the same limit.

Question 9

A road of radius 100 m is banked at 15 degrees, with mu_s = 0.2 between tyres and road. Find the optimum speed and the maximum permissible speed (g = 10 m/s^2, tan 15 = 0.268).

Accepted Answer

Optimum speed, where no friction is needed: v0 = sqrt(Rg tan(theta)) = sqrt(100 x 10 x 0.268) = sqrt(268) = 16.4 m/s. Maximum speed, with friction helping: v_max = sqrt(Rg(mu_s + tan theta)/(1 - mu_s tan theta)) = sqrt(1000 x 0.468/0.946) = sqrt(494) = 22.2 m/s. Check the second formula by putting theta = 0, which correctly gives back sqrt(mu_s R g).

Question 10

A 50 kg person stands on a weighing scale in a lift (g = 10 m/s^2). Find the scale reading when the lift (a) accelerates upward at 2 m/s^2, (b) moves up at constant speed, (c) accelerates downward at 2 m/s^2, and (d) falls freely.

Accepted Answer

The scale reads the normal force N, not the weight. (a) N - mg = ma, so N = m(g + a) = 50 x 12 = 600 N. (b) Constant speed means zero acceleration, so N = mg = 500 N. (c) mg - N = ma, so N = m(g - a) = 50 x 8 = 400 N. (d) In free fall a = g, so N = 0 — this is weightlessness, and the person's mass has not changed at any point.

Practice — Laws of Motion

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