What is the work done by gravitation

[breakoutecho]

A skier lowers a cart 30m down a slope with a sontant speed. The cart has a mass of 90kg. The angle of the slope is 60 degrees. The coefficient of friction is 0.100.
A)Find the work done by friction as the sled moves down the hill.
B) how much work is done by the rope on the sled.
C) What is the work done by gravitation.
D) What is the total work done.

A) I already calculated friction, which was uN with N=mgcos , which gave me -1323J
But I am not sure about the next 3 parts
B) Would work be zero since speed is constant?

C) WG= FgD=mgd= (90kg) (9.8m/s2) is displacement 0m?

D) Would total work be zero, since speed is constant?

 

[ehild]

How is work defined?

ehild

 

[breakoutecho]

Work is Force * distance.

But since speed is constant, does that mean the total work is 0, since there would be no force?
Or would total work still be the addition of the separate works.

 

[ehild]

The total work of all the forces acting on the sled is zero, but the work of the separate forces is not.
Take care when calculating work. It is W=F d cos (theta), theta being the angle between the force F and displacement d.

ehild

 

[rwisz]

A) The work done by friction can be calculated using the formula W = Fd, where F is the force of friction and d is the displacement. In this case, the force of friction is equal to the normal force (N) multiplied by the coefficient of friction (u). So, the work done by friction would be: W = uN * d = (0.100)(mgcos60)(30m) = -1323J.

B) Since the speed is constant, there is no change in kinetic energy and thus no work is done by the rope on the sled.

C) The work done by gravitation can be calculated using the formula W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the change in height. In this case, the change in height is 30m and the work done by gravitation would be: W = (90kg)(9.8m/s^2)(30m) = 26,460J.

D) The total work done would be the sum of the work done by friction and the work done by gravitation. So, the total work done would be: -1323J + 26,460J = 25,137J. This is the net work done on the cart by all forces (friction and gravitation) during its descent down the slope.