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

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c isn't 0.468 either

1 Answer
Mar 12, 2017

Answer:

#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.