Question

What is the magnitude of the frictional force (in N) exerted on a plush toy of
0.125 kg which, when thrown on the floor with a speed of 3.60 m / s, travels 2.20 m
before stopping?

Answer 1

Given

Mass of toy is $m=0.125 kg$

Initial velocity of toy is $v_0=3.60 m/s$

Final velocity is $v=0$

Distance travelled is $d=2.20 m$

The acceleration of toy is

$a=\frac{v^2-v_0^2}{2d}=\frac{0^2-(3.60 m/s)^2}{2(2.20 m)}=-2.945 m/s^2$

According to Newton's second law,

$-f=ma$

So the frictional force is

$f=(0.125 kg)(2.945 m/s^2)=0.37 N$