- #1

Kennedy

- 70

- 2

## Homework Statement

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 man is: (a) –200 J (b) 61 J (c) 140 J (d) 200 J (e) 260 J

## Homework Equations

W = Fdcosθ

F = ma

## The Attempt at a Solution

I started by calculating the force down the slope of the crate due to gravity, which will be sin30(80) = 40 N. Then I determined the net force acting on the crate. Because the acceleration of the crate is -1.5 m/s^2, F = 8.16(-1.5) = -12.24 N. I calculated the mass by taking 80/9.8 = 8.16 kg. Now, I've done this question several times, but I always mess up the signs, but I still don't understand why some quantities are positive and negative. The speed of the crate is decreasing at a rate of -1.5 m/s^2, so the net force should be in the direction opposite to its direction of motion, right? That means that the net force would be acting up the incline. This yields the force of the man to be 12.24 + 40 = 52.24 N. Then, the work done by the man is 52.24(5) = 260 J, but this is wrong. why am I having so many problems with my signs?