A diver of mass 60 kg jumps down on a diving board. The spring beneath the diving board has a spring constant of 5000 N/m.

Question

A diver of mass 60 kg jumps down on a diving board. The spring beneath the diving board has a spring constant of 5000 N/m. The spring compresses a distance of 0.3 m as the board bends down, then it releases and propels the diver upward.

a. How much elastic potential energy is stored in the diving board's spring when it is compressed 0.3 m?

b. How fast is the diver moving when the spring returns to its equilibrium position?

c. How high does the diver go, compared with the height of the unbent diving board?

Answer 1

First, let's write our givens:

m = 60 kg

k = 5000 N/m

Δy = 0.3 m

a) how much elastic potential energy is in the board when it is compressed?

Instead of a diver on a board, think of this a someone jumping off of a platform held up by a spring, with a spring constant given by k.

recall that the equation for elastic potential energy is E = (1/2) * k * Δy²

plugging in the numbers give: E = (1/2) * (5000 N/m) * (0.3 m)² = 225 Nm

sanity check: Nm is equivalent to kg m²/s², which is equivalent to a joule, so our units are correct

c) how fast is the diver moving when the spring returns to equilibrium?

This boils down to an accounting problem- the spring starts with some potential energy (since it starts compressed). When the spring is at equilibrium, it has no potential energy (since it can't do anything)- so where is that energy going? Assuming we're not dealing with any friction, all the potential energy from the spring is transferred to the diver in the form of kinetic energy - K.E. = P.E.

recall that the equation for kinetic energy is: K.E. = (1/2) * m * v²

so after setting K.E. = P.E. : (1/2) * m * v^2 = 225 Nm

after doing some algebra we get: v = sqrt( 2 * 225 Nm / 60 kg) = 2.74 m/s

as before, when you do this, make sure you check your units: N contains a unit of kg, so the kg on the top and bottom cancel, N also has m/s², so multiplying by m gives m²/s², so the sqrt will give units of m/s, which is units of velocity which is what we want

c) how high does the diver go?

Now we need to use one of the kinematic equations

There are usually 4 kinematic equations on a formula sheet:

a) v = v₀ + at

b) Δy = ((v+v₀) / 2)*t

c) Δy = v₀t + (1/2)*a*t²

d) v² = v₀² + 2*a*Δy

consider what variables we're asked for and what variables these equations have- notice that we're never asked about time, so we know we don't need any of the equations that involve time- which are equations a-c

so we're using v² = v₀² + 2*a*Δy

we found the initial velocity (v₀) in the previous section, a is -g, Δy is what we're looking for, so what's the final velocity v? well, when the diver is the highest they're going to go, the diver isn't moving, so at the point we're interested in, the final velocity is 0, so we're left with:

0 = v₀² - 2*g*Δy -> 2*g*Δy = v₀²

the rest is algebra:

Δy = v₀² / (2*g) = (2.74 m/s)² / (2 * 9.81 m/s²) = 0.38 m, and a quick check of the units suggests this is correct.