If a pull of 210 N accelerates a 40-kg child on ice skates at a rate of 5 m/s², what is the frictional force acting on the skates?

If f is the magnitude of friction, then by Newton's second law the net force on the child is

∑ F = 210 N - f = (40 kg) (5 m/s²)

Solve for f :

f = 210 N - (40 kg) (5 m/s²)

f = 210 N - 200 N

f = 10 N

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