How Does Friction Affect Spring Compression on an Inclined Plane?

[nns91]

Homework Statement

A 2kg block is released 4m from a massless spring with a force constant k=100 N/m that is fixed along a frictionless plane inclined at 30 degree.

(a). Find the maximum compression of the spring.
(b) If the plane is rough rather than frictionless, and the coefficient of plane kinetic friction between the plane and the block is 0.2, find the maximum compression.
(c) For the rough plane, how far up the inline will the block travel after leaving the spring ?

2. A 50 000 kg barge is pulled along a canal at a constant speed of 3km/h by a heavy tractor. The townrope makes an angle of 18 degree with the velocity vector or the barge. The tension in the townrope is 1200 N. If the townrope breaks, how far will the barge move before coming to rest ? Assume that the drag force between the barge and water is independent of velocity.

Homework Equations

E= E_mech + E_therm

The Attempt at a Solution

a. I solved it right

b. The friction kinda confuses me. I tried to used conservation of energy.

Initial: E= mgh

Final: E= mg(h-(4+x)sin30) + (1/2)kx^2 + 0.2*mgcos30*(4+x)

then I set them equal to each other and solve for x. However, I got a wrong answer. Where did I do wrong ?2. I have no clue since I cannot picture the problem

 

[LowlyPion]

Conceptually I think you understand the issues involved. The friction diminishes the potential energy available to go into kinetic and then vice versa.

One thing I notice is that your equation for friction is not taking into account the x to detent and then return to equilibrium which is really 2x not x.

For instance:

E= mg(h-(4+x)sin30) + (1/2)kx^2 + 0.2*mgcos30*(4+x)

For the equation at the moment of maximum detent this would be
mg(4+x)sin30 = (1/2)kx² + 0.2mgCos30*(4+x)

Then your return to max height up the slope would be

mg(h+x)sin30 = 1/2kx² - .2mgCos30*(h+x)

2) As regards your barge, note that it is traveling at constant velocity at the moment of snapping. The projection of the 1200N force at the 18° angle that the rope makes with the barge is the forward motive force that, when removed, becomes a decelerating force.

Since F = m*a and you now know a, I'm sure you know how to solve the kinematics associated with coming to a stop.

 

[nns91]

Thanks.

For the 2nd problem:

I found a= -Tension*cos18/m = -0.228 m/s^2

Then I use vf^2= vi^3+2ax and solve for x. Is that true ??

 

[LowlyPion]

Thanks.

For the 2nd problem:

I found a= -Tension*cos18/m = -0.228 m/s^2

Then I use vf^2= vi^3+2ax and solve for x. Is that true ??

That should do it.

I see your typo, so to make it clear:

V² = 2*a*x