Power due to the force applied to the block by the cable?

[ViewtifulBeau]

A 2600 kg block of granite is pulled up an incline which is 36.87 degrees off of the horizontal at a constant speed of 1.07m/s by a cable and winch. The coefficient of kinetic friction between the block and the incline is 0.25. What is the power due to the force applied to the block by the cable?

I started this problem by finding the acceleration of the block which is sin(36.87) *9.81=5.88599m/s^2

This times the mass is the force pulling up. = 15303 N

Next i found the frictional force which is N * u
so cos(36.87) * 9.81 * 2600 * .25 = 5101 N

I subtracted the force pulling up by the frictional force and multiplied it by the velocity 15303-5101 = 10201 * 1.07 = 10916 W

where did i go wrong because this is not the answer. thanks

 

[nevetsnosaj]

I think you mistake was the value of acceleration. When that block move with constant speed, the acceleration = 0.

Once you figured out the some of the forces, you can plug you number into the equation to find your work. It is Force multiplied by distance. In this case, you can assume that this system working in 1 second, so you can get the distance is equal to 1.07m.
Then plug in the W divided by the differences in time to get power. The time is 1 second because we assumed that this things happened in 1 sec.

Is the answer 21810.9 Joule/s? That's what i got.

 

[quicknote]

Your approach is correct, but there is a small mistake in your calculation. The frictional force should be calculated as N * u * cos(36.87), not just cos(36.87). This is because the normal force (N) is acting perpendicular to the incline, while the frictional force is acting parallel to the incline. So the correct calculation would be cos(36.87) * 9.81 * 2600 * 0.25 = 3575 N.

Subtracting this from the force pulling up, we get 15303 - 3575 = 11728 N. Multiplying this by the velocity of 1.07 m/s, we get a power of 12545 W, which is the correct answer.

In the future, make sure to pay attention to the direction of the forces and angles when calculating forces and vectors. Good job on approaching the problem correctly!