A shot-putter moves his arm and the 7.0-kg shot through a distance of 1.0 m, giving the shot a velocity of 10 m/s from rest. Find the average force exerted on the shot during this time.

F = 350 N

Explanation:

Let us consider uniform acceleration,

m = 7 kg

Distance moved, s = 1 m

Final velocity, v = 10 m/s

Initial velocity, u = 0 m/s

Find acceleration (Use Newton's third law of motion),

v^2 = u^2 + 2as

10^2 = 0 + 2*a*1

100 = 2a

a = 50 m/s^2

Find force,

F = ma = 7*50

F = 350 N