Physics problem???? I have two problems that I cannot get. Please Help?

Question

1. A 0.50 kg block, starting at rest, slides down a 30.0° incline with kinetic friction coefficient 0.24 (the figure below). After sliding 82 cm down the incline, it slides across a frictionless horizontal surface and encounters a spring (k = 38 N/m). What is the maximum compression of the spring? and After the compression of part (a), the spring rebounds and shoots the block back up the incline. How far along the incline does the block travel before coming to rest?

2.A crate of mass m1 on a frictionless inclined plane is attached to another crate of mass m2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspended in air. The plane is inclined at an angle = 36.9°. Use conservation of energy to find how fast crate m2 is moving after m1 has traveled a distance of 1.4 m along the incline, starting from rest. The mass of m1 is 13.4 kg and the mass of m2 is 18.8 kg.

Any kind of help would be really really appreciated. Thanks

Answer 1

Lovely energy problems

I'll provide the method, you can plug in the numbers

You need to figure out the velocity of the block at the bottom of the slope
1)
PEi = mgh
h = -.82*sin(30)
PEf = 0
KEi = 0
KEf = 1/2mv^2

The only work done on the system is by friction, therefore

Work = -friction*d = Sum of the Changes in Energy
-m*g*u = PEf - PEi + KEf - KEi
-m*g*u = 0 - mgh + 1/2mv^2 - 0
-g*u = -g*h + 1/2v^2
1/2v^2 = g*(h-u)
v = sqrt(2*g*(h-u))

The kinetic energy of the mass is
KE = 1/2mv^2
That energy gets transferred into the spring's potential energy
PE(spring) = 1/2kx^2

1/2 mv^2 = 1/2 kx^2

x = sqrt(mv^2/k)

For the second part you can just use the velocity you found
Going back up the slope
Work = -f*d = PEf - PEi + KEf - KEi
In this case our PEi = 0 and KEf = 0
-m*g*u = m*g*h - 1/2 mv^2
-g*u = g*h -1/2v^2
gh = 1/2v^2-g*u
h = 1/(2g)*v^2 - u

presto!

2) A FBD of m2 tells us the following
Sum of the vertical forces = m2*a = T - m2*g
T is the tension in the rope
T = m2*(a+g)

A FBD of m1 tells us the following
Sum of the forces along the slope = m1*a = m1*g*sin(theta) - T
T = m1*a - m1*g*sin(theta)

Set the T's equal to eachother and solve for a
m1*a - m1*g*sin(theta) = m2*(a+g)

a*(m1-m2) = m1*g*sin(theta) + m2*g
a = (m1*g*sin(theta) + m2*g)/(m1-m2)

you can use kinematics for the rest...

I got to run, but I'll try to explain part 2 using energy later.