Question

a) A 60 kg person jumps from a platform onto a trampoline 10 m below, stretching it 1.0 m from its initial position. Assuming that the trampoline behaves like a simple elastic spring, how much will it stretch if the same person jumps from a height of 20 m?

b) A baseball of mass 250 g, pitched with a speed of 40 m/s, is caught by the catcher, whose glove moves backward 0.25 m while stopping the ball.
    (i)    What was the kinetic energy of the ball?
        (ii)    How much work did the catcher's glove do on the ball?
        (iii)    What average stopping force was exerted on the ball?

Answer 1

Part A)

First we need to find the spring constant(K), by applying energy conservation we can find it

To find the streching when the person jumps from 20m height, we need to apply energy conservation

Part B)

(i) K.E = (1/2)*m*V² = 0.5*0.25*40*40 = 200 J.

(ii) Work done by the catcher's glove = K.E_f - K.E_i = 0 - 200 = -200 J.

(iii) For average stopping force, we need to apply the formula of work done

Work done = Avg Force*backward distance

-200 = F_avg*0.25

F_avg = -800 N.   -ve sign indicate that the force is in opposite direction of the pitched ball.

Thank You !!!