Question:

A 2 kg block is initially at rest on a frictionless horizontal surface. A constant force of magnitude F = 10 N is then applied to the block in the positive x-direction, parallel to the surface. The block moves a distance of 4 m and comes to a stop. Calculate the work done by the applied force.

Answer:

To calculate the work done by the applied force, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, since the block starts from rest and comes to a stop, its change in kinetic energy is equal to its initial kinetic energy.

Step 1: Calculate the initial kinetic energy of the block: The initial kinetic energy (K_i) is given by the equation: K_i = (1/2)mv^2, where m is the mass of the block and v is its initial velocity. Since the block starts from rest, its initial velocity is 0.

K_i = (1/2)(2 kg)(0 m/s)^2 K_i = 0 J

Step 2: Calculate the final kinetic energy of the block: The final kinetic energy (K_f) is also given by the equation: K_f = (1/2)mv^2, where m is the mass of the block and v is its final velocity. Since the block comes to a stop, its final velocity is 0.

K_f = (1/2)(2 kg)(0 m/s)^2 K_f = 0 J

Step 3: Calculate the work done by the applied force: According to the work-energy theorem, the work done is equal to the change in kinetic energy, which is equal to the initial kinetic energy.

Work done = K_f - K_i Work done = 0 J - 0 J Work done = 0 J

Therefore, the work done by the applied force is 0 J. Since the block is initially at rest and comes to a stop under the action of the applied force, no work is done on the block.

Note: It is important to remember that in this case, work is defined as the work done by the applied force. There might be other forces acting on the block, but since the surface is frictionless and no other force is mentioned, we only consider the work done by the applied force.