A 20-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 13 J. How much work is done by the frictional force on the block as it moves the 2.0 cm?
A) −2.5 J
B) −1.4 J
C) −3.0 J
D) −1.9 J
E) −14 J

Question

Quizplus · Accepted Answer

The work done by the frictional force can be found using the work-energy principle. The initial potential energy stored in the spring when the block is pulled 10 cm is (1/2)kx^2 = (1/2)(8000 N/m)(0.1 m)^2 = 40 J. When the block has moved 2.0 cm toward equilibrium, the potential energy is (1/2)(8000 N/m)(0.08 m)^2 = 25.6 J. The kinetic energy at this point is given as 13 J. The work done by friction is the change in mechanical energy: initial potential energy - (final potential energy + kinetic energy) = 40 J - (25.6 J + 13 J) = 1.4 J. Since friction does negative work, the work done by friction is -1.4 J.

A 20-Kg Block on a Horizontal Surface Is Attached to a Light