Question

An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform down a maximum distance of 0.240 m below its initial position, and then it rebounds. The platform and springs have negligible mass. What is the man's speed at the instant the depresses the platform 0.120m? If the man just steps gently onto the platform, what maximum distance would he push it down?

Answer 1

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

$E_p = E_e$

$mgh = kx^2/2$

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

$80*9.81*2.74 = k0.24^2/2$

$2150.352 = 0.0288k$

$k = 74665 N/m$

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

$E_p = E_{e2} + E_k + E_{p2}$

$mgh = kx_2^2 + mv^2/2 + mgh_2$

$2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12$

$2150.352 = 537.588 + 40v^2 + 94.176$

$40v^2 = 1518.588$

$v^2 = 37.9647$

$v = \sqrt{37.9647} = 6.16 m/s$

When he steps gently, then his gravity force would equal to his spring force

$mg = kx_3$

$x_3 = mg/k = 80*9.81/74665 = 0.0105m$