What is the distance the block moves up the incline before coming to rest?

AI Thread Summary
A block with a mass of 13.0 kg slides down an incline at a 32.8° angle with constant velocity and is then projected up the incline at an initial speed of 3.85 m/s. To determine how far the block moves up before coming to rest, kinetic energy must be calculated, and energy conservation principles applied, accounting for friction and gravitational forces. The coefficient of friction can be derived from the balance of forces during the downward motion, where the force of gravity equals the force of friction. By setting up the equations for kinetic energy and the work done against friction and gravity, the distance the block travels up the incline can be solved. This approach effectively combines energy conservation with friction analysis to find the required distance.
koomba
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Hey guys thanks for the help on the last problem, I just had one more that I was trying to get finished studying for this big exam coming up:

A block with mass m = 13.0 kg slides down an inclined plane of slope angle 32.8 ° with a constant velocity. It is then projected up the same plane with an initial speed 3.85 m/s. How far up the incline will the block move before coming to rest?

Now I know its kind of like a normal inclined plane problem, but the whole pushing it back up part has got me confused. I actually can't even get my first values. :confused: So I probably need help with those too. Man I am stupid :frown: Hope someone can help me! :bugeye: Thanks!
 
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Well, O.K.

You have the mass, you have the initial speed. So calculate the kinetic energy.
After that, when you try to use the energy conservation law, you see that some amount of energy is being taken up by the friction on the plane and the rest by going against gravity.
Now, the question is, how much is being consumed by friction. Well, from the first part, you can calculate the co-effiecient ( say, k) of friction as follows.

For first case,
force of gravity = force of friction (because constant velocity )
so, mg*sin(32.8deg.)=kmg*cos(32.8deg.)
so, k = tan(32.8deg)

now, Second case (let the required distance be x )

Initial K.E. = work by friction + work by gravity

work by friction is kmg*cos(32.8deg.)*x
work by gravity is mg*sin(32.8deg)*x

You know the K.E
Simply solve and find x - the answer!


vibhav
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Thanks! :biggrin:
 
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