Question

A slithering snake travels once around a circle of radius 3.20m. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 80.0N. How much work does the snake do against friction?

Answer 1

The magnitude of the work done by the snake against friction is 402 J.

Step-by-step explanation:

The work is given by:

`W = F*d`

Where:

F: is the force

d: is the distance

The distance traveled by the snake is given by the perimeter of the circle:

`P = 2\pi r = 2\pi 3.20 m = 20.1 m`

Now, the force in the direction of the movement is:

`|\Sigma F| = F_(\mu) = \mu N = \mu W = 0.25*80.0 N = 20 N`

Finally, the work is:

`W = F*d = 20 N*20.1 m = 402 J`

Therefore, the magnitude of the work done by the snake against friction is 402 J.

I hope it helps you!