Find the net work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 54.0 N.

The net work done by friction on the body of the snake is 333.35J

Explanation:

Work done is given by

W = F × s

Where W is the Work done

F is the force

and s is the distance covered

Since we are to determine the work done by friction, then we will determine the frictional force. The frictional force is given by

f = μN = μw

Where μ is the coefficient of friction

N is the normal reaction

and w is the weight

But, F = f

∴ W = μws

From the question

μ = 0.25

w = 54.0 N

Now, we will determine s

From the question,

We are to determine the work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius.

The distance s here is given by the circumference of the circle. Circumference of a circle is given by 2πr

∴ s = 2πr

s = 2 × π × 3.93

s = 7.86π m

Hence,

W = 0.25 × 54.0 × 7.86π

W = 333.35 J

Hence, the net work done by friction on the body of the snake is 333.35J.

the answer would be c