A 263 g ball of clay falls onto a vertical spring.

The spring constant k = 2.52 N/cm.

The clay sticks to the spring and as the spring compresses a distance of 11.8 cm the clay momentarily comes to rest.
Determine the speed of the clay just before it hit the spring in [m/s]?

Respuesta :

As the spring is compressed, it performs

-1/2 (252 N/m) (0.118 m)² ≈ -1.75 J

of work on the clay, while gravity does

(0.263 kg) (9.80 m/s²) (0.118 m) ≈ 0.304 J

on it. Then the total work W performed on the clay ball is approximately -1.45 J.

By the work-energy theorem, W is equal to the change in the clay ball's kinetic energy ∆K. If v is the speed with which it hits the spring, then

W = ∆K

-1.45 J = 0 - 1/2 mv²

Solve for v :

-1.45 J = -1/2 (0.263 kg) v²

v² ≈ 11.0 m²/s²

v ≈ 3.32 m/s