physics(work done)

[匿名]

2013-09-29 5:41 am

A factory worker pushes a 29.5kg crate a distance of 4.5m along a level floor at constant velocity by pushing downward at an angle of 32∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

How much work is done on the crate by friction during this displacement?

How much work is done by gravity?

What is the total work done on the crate?

回答 (1)

用戶4366

2013-09-29 8:04 am

✔ 最佳答案

A force is exert by the worker along 32° below the horizontal, A force of Fcos32∘apply to the crate in horizontal direction, since the crate moves at constant velocity, total force = 0, Fcos32°= frictional force.
Another force Fsin32°acts vertically downward, the normal force is Fsin32°+ mg and frictional force = (Fsin32°+ mg)m

then Fcos32°= (Fsin32°+ mg)m = (Fsin32°+ 29.5x9.8)0.24

Fcos32°= (Fsin32°+ 29.5x9.8)0.24

0.848F = (0.5299F + 289.1)0.24

3.533F = 0.5299F + 289.1

3.0031F = 289.1

F = 96.27 N

frictional force = Fcos32°= 81.64N

displacement = 4.5m

work done = 81.64X4.5 = 367.4 J

Since the crate moves horizontally, no work is done by gravity

What is the total work done on the crate = 367.4 J

2013-09-29 00:07:06 補充：
frictional force = (Fsin32°+ mg)m

The last m should be mu, the coefficient. I don't know why it appears like that.

👍 0 👎 0