Question

100 lb blocks slides down a frictional inclined plane at an angle of 60 degree with the horizontal determine the accelerating force.the time required travel the first 20ft and the time required to travel the next 20 ft

Answer 1

$F= ma$

$F = mg\sin\alpha$

$g=32.17 \frac{ft}{s^2}$

$a=g\sin\alpha$

$s= V_0t+\frac{at^2}{2};V_0=0\text{ as the block starts to slide from rest}$

$t = \sqrt{\frac{2s}{a}}=\sqrt{\frac{2s}{g\sin\alpha}}$

$\text{for the first completed 20 ft}$

$t_1=\sqrt{\frac{2s}{g\sin\alpha}}=\sqrt{\frac{2*20}{32.17*0.87}}=1.2s$

$\text{for the first completed 40 ft}$

$t_2=\sqrt{\frac{2s}{g\sin\alpha}}=\sqrt{\frac{2*40}{32.17*0.87}}=1.7s$

$\text{time of passage of the second segment in 20 ft}$

$t_3= t_2-t_1=1.7-1.2=0.5s$

Answer: 1.7 s - time of passage of the first segment in 20ft;

0.5s - ime of passage of the second segment in 20ft;