Question

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(c)How much work does the gravitational force do on the ball while it is compressing the mattress?

(d)How much work does the mattress do on the ball?

(e)Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses?

(f)What is the value of the spring constant k?

Answer 1

Answer:

(c) 10.29 J

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

(c) The work done by gravitational force is the product of gravity force and the distance compressed

$E_p = mgx = 7*9.8*0.15 = 10.29 J$

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

$E_p = E_e$

$E_e = mgh$

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

$E_e = 7*9.8*1.65 = 113.19 J$

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) $E_e = 113.19 = kx^2/2$

$k0.15^2/2 = 113.19$

$k = 10061 N/m$

Answer 2

Answer:

(C) Wg = 113.2J

(D) Wm = 10.3J

(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.

(F) k = 457N/m.

Explanation:

See attachment below.