Question

Camping equipment weighing 6000 N is pulled across a frozen lake by means of a horizontal rope. The coefficient of kinetic friction is 0.05. The work done by the campers in pulling the equipment 1000 m at constant velocity is:

Answer 1

Answer:

The Work Done is $3 \times 10^5 \ J.$

Explanation:

Since , the equipment is pulled  in horizontal horizontal plane.

Therefore, its normal reaction must be equal to its weight.

So, Normal reaction, $N=6000\ N.$

We know, frictional force , $F=\mu\times N.$

(Where N is Normal reaction and $\mu$ is coefficient of kinetic friction )

Therefore, $F=0.05 \times 6000\ N=300 \ N.$

We know ,

                  $Work Done , W=F\times D$

( Here D is distance covered by the force)

Putting, D=1000 m.

We get , Work Done$=300\times 1000 \ J\\=3\times 10^5 \ J.$