Physics Elena please help!
posted by Anonymous on .
A sled loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope inclined at 40.0o above the horizontal. The sledge moves a distance of 16.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.510 .
(a) What is the tension in the rope?
b) How much work is done by the rope on the sled?
c)What is the magnitude of the force of friction?
(d) What is the mechanical energy lost due to friction?

force pulling sled = T cos 40
force lifting sled = T sin 30
normal force = mg  T sin 40
friction force = 0.51(mg  T sin 40)
acceleration = 0
so
friction force = pulling force
so
0.51(mg  T sin 40) = T cos 40
solve for T
then work = 16 * T cos 40
etc 
Can you tell me the values you are getting so I can make sure I did the algebra right?

How about the other way around? You post the values and I check. I already took the course (in 1955).

For Tension I am getting 117.4

m g = 177 N
.51 (177  .643 T) = .766 T
90.3  .328 T = .766 T
T = 82.5 N
T cos 40 = 63.2 = pull force = friction force
63.2 * 16 = 1012 Joules = work done (also energy lost to friction) 
What is the label for the work? I tried Newtons and that is wrong.

Work is in Joules !!!!

work done against friction = ENERGY lost
That is JOULES ! 
Oh ya, I understand part a and b now, but I am still confused on part c. How would I solve this?

We did part C immediately
T cos 40 = 63.2 = pull force = friction force 
It is not accelerating so friction force is equal and opposite to pulling force.

Now slow down and draw a good picture and follow what I did carefully step by step. I have to go coach a high school sailing team now.