Question

Three clowns move a 345 kg crate 12.5 m to the right across a smooth floor. Take the positive horizontal and vertical directions to be right and up, respectively. Moe pushes to the right with a force of 535 N , Larry pushes to the left with 225 N , and Curly pushes straight down with 705 N . Calculate the work ???? done by each of the clowns. Assume friction is negligible.

Answer 1

Answer:

Moe: 6687.5 J, Larry: -2812.5 J, Curly: 0 J

Explanation:

The work done by each clown is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the box

$\theta$ is the angle between the direction of the force and of the displacement

Let's apply the formula to each clown:

- Moe: F = 535 N, d = 12.5 m, $\theta=0^{\circ}$ (because Moe pushes to the right, and the box also moves to the right)

$W=(535)(12.5)cos 0^{\circ}=6687.5 J$

- Larry: F = 225 N, d = 12.5 m, $\theta=180^{\circ}$ (because Larry pushes to the left, while the box moves to the right)

$W=(225)(12.5)cos 180^{\circ}=-2812.5 J$

- Curly: F = 705 N, d = 12.5 m, $\theta=90^{\circ}$ (because Curly pushes downward, while the box moves to the right)

$W=(705)(12.5)cos 90^{\circ}=0$