A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30 ° with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s², then the work done by the mand is:

Question

Quizplus · Accepted Answer

The force exerted by the man can be resolved into two components: one parallel to the slope and one perpendicular to the slope. Since the slope is frictionless, the only force acting on the crate parallel to the slope is the force exerted by the man, which we'll call F.

The force of gravity acting on the crate can be resolved into two components as well: one perpendicular to the slope and one parallel to the slope. Since the slope is at an angle of 30 degrees with the horizontal, the component of gravity parallel to the slope is given by:

Fgpar = mg * sin30

where m is the mass of the crate, and sin30 is equal to 0.5.

Since the crate is moving upward, the direction of the force of gravity is downward, so the force of gravity is equal in magnitude but opposite in direction to Fgpar:

Fg = -mg * sin30

Using Newton's second law, we can write an equation for the acceleration of the crate:

F - mg * sin30 = ma

where a is the acceleration of the crate.

We know that the speed of the crate decreases at a rate of 1.5 m/s, so we can write:

a = -1.5 m/s^2

Substituting this into the previous equation and solving for F, we get:

F = ma + mg * sin30 = (-1.5 m/s^2)(m) + (9.8 m/s^2)(1 kg)(0.5) = 2.35 N

The work done by the man is equal to the force he exerts multiplied by the distance the crate moves:

W = F*d*cos30 = (2.35 N)(5.0 m)(cos30) = 10.8 J

But we have to be careful with the signs. The force and the displacement are in opposite directions, so the cosine of the angle between them is negative. Therefore:

W = -10.8 J

which to the nearest 10 J gives us choice C as the correct answer.

A Man Pushes an 80-N Crate a Distance of 5