Question

A block of mass $4kg$ slides down a plane inclined at ${37}^{\circ }$ with the horizontal. The length of plane is $3m$. The coefficient of sliding friction between the block and the plane is $0.2$. Find the work done by gravity and the frictional force on the block.

• A. 0J, -18.816 J
• B. -70.56J, 0J
• C. 70.56J , -18.816 J
• D. None of these

Answer 1

The correct option is C

70.56J , -18.816 J

As the normal is perpendicular to the displacement, work done by the normal reaction $R=Rscos{90}^{\circ }=0$

The magnitude of displacement = s = 3m and the angle between force of gravity (mg) and displacement is equal to $({90}^{\circ }-{37}^{\circ })$. Work done by the gravity - $mgscos({90}^{\circ }-{37}^{\circ })$

$\Rightarrow mgsin{37}^{\circ }$ = 4 x 9.8 x 3 x 3/5= 70.56 J

Work done by friction = -$\left(\mu mgcos{37}^{\circ }\right)$ s= - 0.2 x 4 x 9.8 x 4/5 x 3 = -18.816 J