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?

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

Using the kinematic equation v_f² = v_i² + 2ax we can solve for the acceleration:

v_f² - v_i² = 2ax

(v_f² - v_i²)/(2x) = a

Plugging in the values v_f = 0 (the final velocity is zero), v_i = 3.60 and x = 2.20 we get:

a = (0² - 3.60²)/(2•2.20)

a = -12.96/4.40 = -2.94545 m/s²

Since F = ma, and the only force on the object is the frictional force, then F = (0.125)(-2.94545) = -0.368 N

The negative means it is in the direction opposite to the direction of travel.