Question

A force of magnitude 25.0 N acting at a 60.0° angle to the horizontal pulls a block 5.00 m along the floor. a) Find the work done on the box: W= Joules

2)Now consider if the box experiences a frictional force F, = 5,00 N.
b) If the box moved 5.00 m, find the net work done on the box:
W= Joules

Answer 1

The work done on the block is 106.8 Joules. The net work done on the box is 0 Joules.

Step-by-step explanation:

To find the work done on the block, we need to use the formula:

Work (W) = Force (F) x Distance (d) x cos(theta)

where F = 25.0 N, d = 5.00 m, and theta = 60.0°.

Plugging in the values, we get:

W = 25.0 N × 5.00 m × cos(60.0°) = 106.8 J

Therefore, the work done on the block is 106.8 Joules.

Now, considering the frictional force of 5.00 N, the net work done on the block can be calculated as:

Net Work (W) = Work by Applied Force - Work by Friction

Since the box moves 5.00 m, and the angle of frictional force is 180° (opposite direction to the displacement), we have:

Work by Applied Force = Force x Distance x cos(0°) = 25.0 N × 5.00 m × cos(0°) = 125 J

Work by Friction = Force x Distance x cos(180°) = 5.00 N × 5.00 m × cos(180°) = -125 J

Therefore, the net work done on the box is:

Net Work = 125 J - 125 J = 0 J

Thus, the net work done on the box is 0 Joules.