Physics Problem --> Help!

[mAdD INDIAN]

Can you guys/gals please help me with this physics problem??

A 4kg block is given an initial speed of 8m/s at the bottom of a 20degree incline. The frictional force that retards its motion is 15N.

a) If the block is directed up the incline, how far will it move before it stops?
b) Will it slide back down the incline?


How exactly would I go about solving it? I can find the momentum along the x & y components but that won't help me find the force of the block or the work it does.

I found the kinetic energy which also equals work done by the block.

These are the formulas::

Ek = 1/2mv^2
deltaEk = W = Fdcos(theta)

 

[GreenBeret]

Calculate the force on the block parallel to the plane due to gravity. Using 9.8 should be accurate enough constant for that.

Then do some kinematics to see how long that force, combined with friction takes to stop the block.

Then compare the friction to the parallel force due to gravity and see if it slides back down.

 

[Thorsky]

The other way, which you seem to be wanting based on the equations you provided, is to use energy. Let d be the distance the block travels up the incline. The initial energy of the block is all kinetic: 1/2mv^2. As the block travels up the plane, the kinetic energy is converted into potential energy, and also some is being dissipated by the friction. How much energy is lost to friction? Well, it's just the old force time distance formula...you know the force and use d for the distance. The last thing to account for is the end potential energy, which is mgh where h is the height, m the mass, and g the infamous 9.8m/s^2. Since the plane is inclined at 20 degrees, h=d sin(20). (Be sure your calculator is in degree mode and not radians )

The equation to solve is thus:

1/2mv^2=df+mgd sin(20)

You know m (mass), v(initial velocity), f(frictional force), and g(9.8). So, you can solve for d.

As for determining if the block slides back down, you need to see if the force due to gravity along the plane, mg sin(20), is greater than the force f exerted by the friction. This isn't quite correct as f is the dynamic friction force and what's important for this part of the problem is the static friction force, which is larger (or at least not smaller).

 

[mAdD INDIAN]

Thank you Thorsky! It really helped.

As for Greenberet...I have no idea what your talking about!

 

[Triumph]

one of my professors always said, "if you don't know the answer, just say free body diagram!" so i say to you, free body diagram!