Question

A cat of mass 12 kg is at rest. How much work is needed to accelerate the cat to 5.5 m/s? If this work is done over 10.0 m, what average force is needed to accelerate the cat?

Answer 1

The work done is 180 J. The force involved was 18 N

Explanation:

I will assume the cat is moving horizontally. By the principle of conservation of energy, the work done will equal the kinetic energy of the cat when up to speed.

The final KE is given by
`KE = (1/2) * m * v^2 = (1/2) * 12 kg * (5.5 m/s)^2 = 181.5 J`

Therefore the work done is about 180 J (rounding to the proper number of significant digits).

The work done is also the `"force"*"distance"`. Therefore

`181.5 J = F*10.0 m`

`F = (181.5 J)/(10.0 m) = 18.15 J/m`

Since the Joule is equivalent to Newton*meters,

`F = (18.15 J)/m = (18.15 N*m)/m = 18.15 N ~= 18 N`

I hope this helps,
Steve