Question

Consider the 65.0 kg ice skater being pushed by two others shown in the figure?

c isn't 0.468 either

Answer 1

`A = F/m = 30.44/65.0 = 0.468` `ms^{-2}`

That is the correct answer in physics. So much the worse for whoever wrote the question is not the answer they have set!

Explanation:

The initial velocity is irrelevant, since velocity is relative anyway. The net force acting is `30.44` `N` on a mass of `65` `kg`, so the acceleration has to be `0.468` `ms^{-2}`.

The number of significant digits is even correct.

Edit: The reminder about friction is interesting. We are told she is wearing steel-bladed skates on ice. Was there earlier information on frictional coefficients for that situation? If such, there would be a small force opposing the accelerating force, so her acceleration would be less. You could calculate that as follows:

`F_{frict} = \muF_{norm} = \mumg`

You would subtract that frictional force from the applied force before moving to the `a=F/m` step.