Physics, 12.07.2019 17:10 baylabrown

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.

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Answer from: shahidcade2721

The work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.What is work done?

Work done is the force applied on a body to move it over a distance. Work done to move a body by the application of force can be given as,

$W=Fd\cos\theta$

Here (F) is the magnitude of force, $\theta$ is the angle of force applied and (d) is the distance traveled.

The mass of the crate is 345 kg. The crate moved by the clowns is 12.5 m to the right across a smooth floor.

Moe pushes to the right with a force of 535 N-

As, the Moe applies the force to the crate on the right side and the crate move in the direction of force. Here, the angle made is equal to the 0 degrees.

Therefore, the work done by the Moe is,

$W=535\times12.5\times\cos (0)\\W=6687.5\rm J$

thus, the work done by the Moe is 6687.5 J.

Larry pushes to the left with 225 N. -

As, the Larry applies the force to the crate on the left side and the crate move in the opposite direction of force. Here, the angle made is equal to the 180 degrees.

Therefore, the work done by the Larry is,

$W=225\times12.5\times\cos (180)\\W=-2812.5\rm J$

thus, the work done by the Larry is -2812.5 J.

Curly pushes straight down with 705 N.-

As, the Curly applies the force to the crate on the straight down and the crate move in the perpendicular direction of force. Here, the angle made is equal to the 90 degrees.

Therefore, the work done by the Curly is,

$W=705\times12.5\times\cos (90)\\W=0\rm J$

thus, the work done by the Curly is 0 J.

Thus, the work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.

Learn more about the work done here;

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Answer from: bay97

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)

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

Answer from: Quest

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

Answer from: Quest

i think the answer would be reserve.

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