Question

A 55 kg softball player slides into second base, generating 940 J of thermal energy in her legs and the ground. How fast was she running?

Answer 1

The softball player's running speed before sliding into second base is calculated by equating the generated thermal energy to the player's kinetic energy just before the slide and solving for velocity in the kinetic energy equation KE = (1/2)mv².

Step-by-step explanation:

To find out how fast the softball player was running before sliding into second base, we use the concept of conservation of energy, specifically the conversion of kinetic energy to thermal energy. The thermal energy generated during the slide is equal to the kinetic energy the player had due to her motion just before the slide. Therefore, we can use the formula for kinetic energy, KE = (1/2)mv², where m is mass and v is velocity, and solve for v.

Given that the softball player has a mass (m) of 55 kg and generated 940 J of thermal energy, we have:

940 J = (1/2)(55 kg)v²

To solve for v, we rearrange the equation:

v² = (2 * 940 J) / (55 kg)

v = √((2 * 940 J) / (55 kg))

After calculating the above expression, we would get the player's velocity in meters per second (m/s), which would tell us how fast she was running before she started to slide.

Answer 2

When the player slides into the base, all its kinetic energy converts into thermal energy in the legs and the ground.
The thermal energy generated is 940 J, so the initial kinetic energy of the player must be 940 J as well:

`K= (1)/(2)mv^2=940 J`
where m=55 kg is the mass of the player and v is his speed. Re-arranging the fomula, we can find the value of his speed, v:

`v= \sqrt{ (2K)/(m) }= \sqrt{ (2\cdot 940 J)/(55 kg) } =7.4 m/s`