Question

A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver's speed immediately after leaving the raft is 3.8 m/s. If the time interval of the interaction between the diver and the raft is 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?

Answer 1

Answer:

F = 1094.4 N

Explanation:

From impulse - momentum theorem, we now that ;

Impulse = momentum

Where;

Formula for impulse = force (F) × time(t)

Momentum = mass(m) × velocity(v)

Now, we are given;

Mass of swimmer; m = 72 kg

Speed; v = 3.8 m/s

Time; t = 0.25 s

Thus;

F × t = mv

F = mv/t

F = (72 × 3.8)/0.25

F = 1094.4 N

This value of force is the magnitude of the average horizontal force by diver on the raft.