Question

A 100 lb block rest on the floor and the coefficient of sliding friction between block and floor is 0.25. A horizontal force of 40 lb acts on the block for 3s accelerates it. compute the velocity of the block at the end of 3s

Answer 1

Let's first find the acceleration of the block:

$$F_h-F_{fr}=ma,$$ $$F_h-\mu_sW=\dfrac{W}{g}a,$$ $$a=\dfrac{F_h-\mu_sW}{\dfrac{W}{g}},$$ $$a=\dfrac{40\ lb-0.25\times100\ lb}{\dfrac{100\ lb}{32.17\ \dfrac{ft}{s^2}}}=4.82\ \dfrac{ft}{s^2},$$ $$a=4.82\ \dfrac{ft}{s^2}\times\dfrac{0.3048\ m}{1\ ft}=1.47\ \dfrac{m}{s^2}.$$

Finally, we can find the velocity of the block at the end of 3s from the kinematic equation:

$$v=v_0+at=0+1.47\ \dfrac{m}{s^2}\times3\ s=4.41\ \dfrac{m}{s}.$$