Question

A 120N force applied horizontally drags a 4kg load along a horizontal table at a uniform velocity of 3m/s (a) what is the coefficient of kinetic friction between the load and the table (b) if the applied force is suddenly removed,how much further does the load slide before coming to a stop

Answer 1

Explanations & Calculations

a)

• Applied force is equal to the dynamic friction force at any event of uniform speed. Therefore,

$\qquad\qquad \begin{aligned} \small F&=\small f\\ &=\small \mu R=\mu mg\\ \small \mu&=\small \frac{120N}{4kg\times9.8ms^{-2}}\\ &=\small 3.1 \end{aligned}$

b)

• By the time the force is removed the load has some kinetic energy. This is expended during the travel ahead to overcome the friction applied. Therefore,

$\qquad\qquad \begin{aligned} \small E_k&=\small fs\\ \small \frac{1}{2}mv^2&=\small \mu mg.s\\ \small s&=\small \frac{v^2}{2\mu g} \end{aligned}$

• Now it is easy to substitute the values and try getting the answer.