Physics Question: Harold the Hurler

Question

Harold the Hurler is a physics student who is renowned for his baseball pitching arm. In order to determine the amount of work he performs in throwing a baseball, Harold devises an experiment. He stands at the bottom of a deep pit and hurls a 0.141 kg baseball through an open third‑floor window in a nearby building. Harold's aim is so accurate that the ball then smoothly enters the tube of the Baseball Absorber that Harold invented and patented. In this device, the ball compresses a spring until it comes momentarily to rest, and this maximum amount of compression is recorded as 0.419 m. The spring's force constant is 835 N/m, and the position of the baseball's momentary rest is 7.13 m above ground level. The point in the pit where the Hurler starts his pitch is 13.9 m below ground level.

How much work does Harold perform on the baseball? Take g=9.80 m/s^2

Answer 1

Let

m = 0.141 kg be the mass of the ball,

d = 13.9 m be the depth of the pit,

h = 7.13 m be the height of the point where the ball comes to rest,

k = 835 N/m be the spring constant,

s = 0.419 m be the compression of the spring when the ball stops,

g = 9.8 m/s² be gravitational acceleration.

The work W performed by Harold gets converted to energy of the system; that includes

- kinetic energy of the ball,

- potential energy of the ball,

- spring energy,

- heat due to resistance and friction, which I assume we are to ignore.

By conservation of energy the energy of the system remains constant, and at every point

in time equals the work W.

At the moment the ball leaves his hand all the energy is in the form of kinetic energy.

We chose to make potential energy initially 0, by using the bottom of the pit as the reference point for potential energy.

When the ball comes to rest inside the Baseball Absorber,

the ball's kinetic energy is 0 (because its speed is 0),

the ball's potential energy is mg(h+d), because the ball is distance h+d from the reference point for potential energy,

the spring's energy is ks²/2.

Therefore the work is

W = mg(h+d) + ks²/2

Substituting actual numbers

W = 0.141×9.8×(7.13+13.9) + 835×0.419²/2 = 102.4 J