A 90 kilogram man jumps from rest from a height of 2.5 meters onto a platform mo

Question

A 90 kilogram man jumps from rest from a height of 2.5 meters onto a platform mounted on a spring. As the spring compresses, the platform is pushed down a maximum distance of 0.2 meter below its initial position, and then it rebounds. The platform and the spring have negligible masses.

a) What is the spring constant of the spring?

b) What is the man's speed at the instant the platform is compressed 0.1 meters?

c) If the man has stepped gently onto the top of the platform, what maximum distance would it have been pushed down?

Explanation / Answer

The mass of the man, m = 90 kg

Height, h = 2.5 m

The compression in the spring, x = 0.2 m

a) According to the law of conservation of energy

m g h  = (1/2) k x^2

k = (2mgh) / x^2 = 110250 N/m

b) The compression in the spring, x = 0.1 m

 Acoording to law of conservation of energy, we have

(1/2) m v^2 = (1/2) k x^2

 v^2 = [k x^2]/m = 12.25

v = 3.5 m/s

c) If the man stepped gently, then

mg = kx

x = mg / k = 0.008 m

So the compression in the spring is given by, x = 0.008 m = 0.8 cm

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