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.
A) -330 J
B) 0 J
C) -3300 J
D) -670 J

Question

Quizplus · Accepted Answer

The net work done by friction is negative because the force of friction opposes the motion of the snake. The magnitude of the work done by friction is equal to the product of the force of friction and the distance traveled by the snake. The force of friction is equal to the coefficient of friction times the normal force, which is equal to the weight of the snake. The distance traveled by the snake is equal to the circumference of the circle, which is 2πr. Therefore, the net work done by friction is:$$W_f = -f*d = -μ*N*2πr = -0.25*54.0 N*2π*3.93 m = -330 J$$

Find the Net Work Done by Friction on the Body