Question

A hand pushes a 2 kg block along a table from point A to point C as shown in the figure below. The table has been prepared so that the left half of the table (from A to B) is frictionless. The right half (from B to C) has a non-zero coefficient of friction equal to . The hand pushes the block the full distance from A to C using a constant force of 5 N. The block starts off at rest at point A and comes to a stop when it reaches point C. The distance from A to B is 0.5 m and the distance from B to C is also 0.5 m. We will consider the block by itself as our system so all forces on the block are external A. From A to B: The work done by the weight force is The work done by the normal force is The work done by the pushing force is The change in kinetic energy of the block is . B. From B to C: The work done by the weight force is The work done by the normal force is The work done by the pushing force is . The work done by the friction force is - - The change in kinetic energy of the block is so The magnitude (absolute value) of the work done by the friction force is Se the magnitude (absolute value) of the work done by the pushing force. C. From A to C: The change in kinetic energy of the block is so D. From A to B: What is the work done by the pushing force? What is the change in kinetic energy of the block? What is the speed of the block at point B? m/s E. From B to C: What is the work done by the pushing force? What is the change in kinetic energy of the block? What is the work done by the friction force? What is the coefficient of friction?

Answer 1

W = F ds

Work-Energy theorem

Wietal = AK.E.

f friction = UN

By These three above mentioned equation.

A.

0, 0, +2.5J and +2.5J

B.

0, 0, +2.5 J, -5 J, -2.5 J, Double

C.

0

D.

+2.5 J, +2.5 J

E.

+2.5 J, -2.5 J, -5 J and 0.5