Question

A body of mass (M) 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is

Answer 1

$N_1=Mg\cos {45}^{\circ }$
$\therefore Mg\sin {45}^{\circ }-\mu N_1=Ma$
$Mg\sin {45}^{\circ }-\mu Mg\cos {45}^{\circ }=Ma$
$a=g\sin {45}^{\circ }-\mu g\cos {45}^{\circ }$
$=3.468\frac{m}{s^2}$
Now, from 2nd equation of motion.
$V^2=u^2+2as$
$(20{)}^2=0+2\times 3.468\times s$
$s=57.667m$