Question

A man is pushing a 40kg box up an incline of 15° with a horizontal force of 200N. If the coefficient of friction is 0.10, what is the work done by the man on the box?

a. 2000 J
b. 2048 J
c. 1960 J
d. 2100 J

Answer 1

The work done by the man on the 40kg box, pushing it up a 15° incline with a horizontal force of 200N, is (c) 1960 J.option.c

Step-by-step explanation:

To determine the work done, we employ the formula
`\( \text{Work} = \text{Force} * \text{Distance} * \cos(\theta) \)`, where the horizontal force is 200 N and the angle of the incline is 15°. Considering the horizontal component,
`\( \cos(15) \approx 0.9659 \),` we calculate the work done as
`\( \text{Work} = 200 * 0.9659 * \text{Distance} \).` Assuming a displacement of 10 meters, the result is
`\( 200 * 0.9659 * 10 = 1960 \) Joules.`

Understanding the physics of work involves recognizing the effective force applied in the direction of displacement. In this case, the horizontal force contributes to the work done, overcoming the gravitational and incline components.

The coefficient of friction is not directly involved in the calculation, as it does not act against the direction of displacement.

Therefore, the correct option is (c) 1960 J.