What is the kinetic energy of the block?

[anyone1979]

[SOLVED] Slide, I

I know I messed up on this problem somewhere, Please anybody help me figure out where.

A block of mass m = 1 kg slides down an inclined plane 2 m high and 4 m long(measured alond the incline). The coefficient of kinetic friction is uk = .07.
(a) What is the kinetic energy of the block when it reaches the bottom of the plane?
(b) The block continues to slide over a level plane with the same value of uk. How far does it go before coming to rest?

a) a = gsin(theta)

theta = arcsin(2/4) = 30 degress
a = (9.8)sin(30) = 4.9 m/s^2

Vf^2 = 0 + 2(4.9)(4)
Vf = 6.3 m/s

KE = 1/2(mVf^2)
KE = 1/2(1)(6.3)^2
KE = 19.8J

I can not figure out how to work in kinetic friction nor part b.

 

[hotcommodity]

The friction is a non-conservative force, and since the frictional force points opposite to the motion of the block, it will do negative work on it. The relationship of the net work done by non-conservative forces to the objects mechanical energy is W_{nc} = E_f - E_0, and so your equation will look like:

W_{nc} + mgh_0 = .5mv^{2}_{f}

Assuming that we take the bottom of the slide to have zero potential energy, and assuming the block accelerates from rest. Does that help?

 

[anyone1979]

This is what I got for the friction:

fk = uk(N)
N = mg sin(30) = 4.9N
fk = (0.07)(4.9) = .343N

Wfk = fk(x)cos(180)
Wfk = .343(4)(-1) = -1.372J (work done by friction)

How do I combine the work done by friction and kinetic energy together?
Also how do I even begin part b? Thanks

 

[hotcommodity]

I believe there is a mistake in calculating your normal force. You should be using the cosine instead of sine. Other than that, you've got the right idea. The work done by friction, in this case, is the work done by all non-conservative forces. This is how we relate the work done by non-conservative forces to an objects change in mechanical energy:

W_{nc} = E_f - E_0 = [.5mv^{2}_{f} + mgh_f] - [.5mv^{2}_{0} + mgh_0]

The block has zero potential energy at the bottom of the incline, and initially has zero kinetic energy, as it accelerates from rest, so we have:

W_{nc} = .5mv^{2}_{f} - mgh_0

Solve for the final kinetic energy:

.5mv^{2}_{f} = K_f = W_{nc} + mgh_0

You know how to find the work done by friction, and now you can just plug in numbers.

For part b, you want to use the fact that while the block slides along the floor, it has a constant deceleration. Use Newton's Second Law to find this deceleration, f_k = ma, and use the equations of kinematics to find the distance traveled.

 

[anyone1979]

Thank you once again.

So, Wnet = Wgravity + Wfk?
Both of them are negative, but since work cannot be negative, I just add them?

Thanks, I solve part b by finding the normal formal once it on on the level plane, found the kinetic friction, the accerlation, used the velocity 6.3 m/s as my initial and 0 as my final velocity, found the time and solved for x.

 

[hotcommodity]

Well let's be careful about what we're saying. Work can be negative. Work is defined as an objects change in kinetic energy. If I throw a box such that it slides along the ground, and the only force acting along its path of motion is friction, that friction will decrease the box's kinetic energy, and therefore negative work will be done on the box because it had a higher amount of kinetic energy when it first began to slide than it did when it finally came to rest. Does that make sense?

When the box is sliding down the incline, the x component of the weight vector does positive work on the box because it causes it to accelerate from rest, and so it has an increasing amount of kinetic energy, therefore its net work is positive. So you shouldn't get a negative value for the work done by the gravitational force. You can use the equation I gave you above, but in general, the net work done on an object is the sum of the work done on the object by conservative forces and the work done on the object by non-conservative forces: W_{net} = W_{nc} + W_c

By the work energy theorem, we also know that the total work done on an object by a constant force is equal to the objects change in kinetic energy:

W_{net} = \Delta K

I hope that helps.

 

[anyone1979]

Yes it makes more sense now...Thank you