Question

1. A force F is exerted on a 2.0 kg block to move it across a rough surface. The

magnitude of the force is initially 5 N, and the block moves at constant velocity.
While the block is moving, the force is instantaneous increases to 12 N. How much
kinetic energy does the block now gain at it moves a distance of 2 m?

Answer 1

The block gains 10 J of kinetic energy as it moves a distance of 2 m.

Step-by-step explanation:

First, recognize that the initial force acting on the block is 5 N and the final force is 12 N. The block moves at constant velocity, so the net force must be zero. From this, we can determine that the force of friction acting on the block is equal to the applied force of 5 N.

The work done by a force is given by the formula W = Fs, where W is the work done, F is the force, and s is the distance moved. In this case, the force of friction is 5 N, and the distance moved is 2 m. Therefore, the work done by friction on the block is 10 J.

The work done by a force is also equal to the change in kinetic energy. Since the block starts at rest and ends at a constant velocity, the work done by friction is equal to the kinetic energy gained by the block. Therefore, the block gains 10 J of kinetic energy as it moves a distance of 2 m.