Question

A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. Assume constant acceleration of the hammer-nail pair.

a) 1.0 x 10^2 N
b) 1.5 x 10^2 N
c) 2.0 x 10^2 N
d) 2.5 x 10^2 N

Answer 1

The average force exerted on the nail by the 0.450-kg hammer moving at 7.00 m/s and coming to rest after driving the nail 1.00 cm into a board is approximately 1.1 x 10³ N.

Step-by-step explanation:

Given a 0.450-kg hammer moving at 7.00 m/s comes to rest after driving a nail into a board, we can calculate the average force exerted on the nail using the work-energy principle. Since work done is equal to the change in kinetic energy and the hammer comes to a stop, the kinetic energy of the hammer is completely used to do the work of driving the nail.

First, we calculate the initial kinetic energy (KE) of the hammer:

KE = (1/2)mv² = (1/2)(0.450 kg)(7.00 m/s)² = 10.95 J

Then, assuming the nail is driven 1.00 cm (0.01 m) into the board, the work (W) done by the hammer on the nail is:

W = KE = 10.95 J

Therefore, the average force (F) exerted on the nail is:

F = W/d = 10.95 J / 0.01 m = 1095 N

Thus, the average force exerted on the nail is 1095 N, which can be approximated to 1.1 x 10³ N due to significant figures.