http://www.smartphysics.com/images/content/mechanics/ch9/blockonincline.png

A mass m = 12 kg is pulled along a horizontal floor with NO friction for a distance d =5.2 m. Then the mass is pulled up an incline that makes an angle θ = 35° with the horizontal and has a coefficient of kinetic friction μk = 0.34. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 35° (thus on the incline it is parallel to the surface) and has a tension T =55 N. The speed of the block right before it begins to travel up the incline is 6.249 m/s.

How far up the incline does the block travel before coming to rest? (Measured along the incline.)
I got 5.17 m, but I am not sure if I am right! If I did wrong, please tell me how I should do it!


On the incline the net work done on the block is:
1. positive
2. negative
3. zero

it think it should be negative because the work done by tension is positive, but the magnitude of work done by friction and gravity is negative and is much bigger in magnitude, so the net work done should be negative, right?

Along the incline, the forces that apply are friction, weight and the pulling force.

\mu R + mg \sin\theta - F = ma

taking g = 9.8 m/s^2

(0.34)(12\times9.8) + (12\times9.8)(\sin 35) - 55 = 12a

I get a = 4.37 m/s^2

So, distance up is:

v^2 = u^2 + 2as

I get

s = \frac{-(6.249^2)}{2(-4.37)} = 4.60\ m

For the second part, remember that work done is equal to force x displacement. A force was applied and the mass moved a certain distance... so?

There is a simpler way to solve this problem without having to rely on the equations of motion. Remember:
1. W[net] = Force[net] * displacement
2. W[net] = delta K = (1/2)m*v^2 - (1/2)m*vi^2 where vi = initial velocity

Using equation 2, (1/2)m*v^2 = 0 since v^2 will be 0 once the object comes to rest.
Thus, W[net] = -(1/2)m*vi(2)

Substituting this into equation 1, we get:
W[net]/F[net] = displacement
where F[net] = F[kinetic friction] + F[weight] - F[Tension] = u[k]*g*Cos[theta] + m*g*Sin[theta]-T
==> displacement = -(1/2)m*vi^2 / ( u[k]*g*Cos[theta] + m*g*Sin[theta]-T )

You can plug in the numbers yourselves, the equation is now in terms of all known variables.

Having done the calculations, you will see that the answer matches that of the answer above me. This is just another solution, relying solely on the equations of Work.

Indeed, but I like to use both independently, to make sure my answers are correct.

I was actually planning to show the asker the other method too once (s)he gets to understand how to solve the problem, but since (s)he never got back :(