Question

9.) A 19.6 kg block is dragged over a rough, horizontal surface by a constant force of 183 N acting at an angle of angle 30.5 ◦ above the horizontal. The block is displaced 97.1 m and the coefficient of kinetic friction is 0.121.

a. Find the work done by the normal force. Answer in units of J.

b. What is the net work done on the block? Answer in units of J.

Answer 1

Mass of block=19.6kg

Force=183N=F

Angle between force and displacement =30.5°=X

Distance moved by block=S=97.1m

As force is constant, work done by the force =FSCosx

Work done =183*97.1*cos(30.5)=17769.3*0.861=15310.28j

As force was acting in backward direction and displacement was in forward direction so angle between displacement and force =180° and cos180=-1

Force=ceoff.of kinetic friction * normal

As body was makes an angle of 30.5° with horizontal we have

Normal+Fsinx=weight(mg)

Normal+(183*0.507)=19.6*9.8

Normal+92.88=192.08

Normal=99.2N

Coefficient of kinetic friction =0.121

So frictional force=0.121*99.2=12.00N

Work done by friction=FSCosx

12*97.1*-1=-1165. 51j

Now total work done scalarly=16475

This will be the work done by force F and the work done by the friction. As both are works done on the same body, so we add up them to get the total work.

If it was asked keeping directions in mind then net work done will be=14145 j