A 3.00kg block is released from rest at the elevated end of A (top of a ramp) of a frictionless track that slopes down to a point B and continues horizontally to a point D, beyond which is a rough horizontal surface (friction). The speed of the block at a point C between B and D (end of the plane and beginning of surface with friction) is 7.2m/s, and the block comes to rest after sliding 2.77m along the rough surface.
Find the work done on the block by the normal force exerted by the track.
I got 0.
Find the work done on the block by the gravitation force by earth.
I got 30dsin(theta). Can I find theta or d?
Find the work done on the block by the frictional force exerted by the rough surface.
I got Wf = muk*Fn*2.77m
I don't know what to do with normal force or the coefficient of kinetic friction in this one...
Find the coefficient of kinetic friction between the block and the rough surface.
I have no idea what to do for this one.
Can someone check my answers and help me out on the parts I couldn't get!
Find the work done on the block by the normal force exerted by the track.
I got 0.
Find the work done on the block by the gravitation force by earth.
I got 30dsin(theta). Can I find theta or d?
Find the work done on the block by the frictional force exerted by the rough surface.
I got Wf = muk*Fn*2.77m
I don't know what to do with normal force or the coefficient of kinetic friction in this one...
Find the coefficient of kinetic friction between the block and the rough surface.
I have no idea what to do for this one.
Can someone check my answers and help me out on the parts I couldn't get!
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1. The work done by a normal force is zero. Correct.
2. You are correct but don't have this information. Use the fact that energy is conserved.
Ek = 1/2 m v^2
= 1.5 * 7.2^2
=77.8J This must have come from gravity and, as it is prior to the friction, it must exactly equal the work done.
3. As it is now at rest, the work done must equal the energy in the block so it must equal the result of 2.
Your answer is once more correct but you use it AFTER you have found the energy as in my answer 2 and 3.
4.You may now substitute E = 77.8, Fn= 3.00 * 9.8, d = 2.77 m
so 77.8 = muk * ( 3.00 * 9.8) * 2.77
muk = 77.8 /( 3.00 * 9.8 * 2.77)
=0.96
2. You are correct but don't have this information. Use the fact that energy is conserved.
Ek = 1/2 m v^2
= 1.5 * 7.2^2
=77.8J This must have come from gravity and, as it is prior to the friction, it must exactly equal the work done.
3. As it is now at rest, the work done must equal the energy in the block so it must equal the result of 2.
Your answer is once more correct but you use it AFTER you have found the energy as in my answer 2 and 3.
4.You may now substitute E = 77.8, Fn= 3.00 * 9.8, d = 2.77 m
so 77.8 = muk * ( 3.00 * 9.8) * 2.77
muk = 77.8 /( 3.00 * 9.8 * 2.77)
=0.96
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if the block started from rest, it converted all its PE into KE at the bottom of the ramp, and you know its speed at the bottom of the ramp is 7.2m/s since that is its speed on the flat portion before any friction sets in
so, use the work energy theorem to relate work done by gravity to the change in KE
at the top, the block had no KE, at the bottom, its KE is 1/2 x 3kg x (7.2m/s)^2...this is the work done by gravity
since the friction surface slows this block to zero, the work done by friction is just the negative of this
so, use the work energy theorem to relate work done by gravity to the change in KE
at the top, the block had no KE, at the bottom, its KE is 1/2 x 3kg x (7.2m/s)^2...this is the work done by gravity
since the friction surface slows this block to zero, the work done by friction is just the negative of this