Answer to Question #279061 in Mechanics | Relativity for Zizu

Question

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

a. if the plane is frictionless, find the maximum compression of the spring.

b. if the coefficient of kinetic friction between the plane and the block is 0.2, find the maximum compression.

c. for the plane in part(b), how far up the incline will the block travel after leaving the spring?

Answer 1

Solution:

A) Energy conversion between gravitational and elastic energy:

$$\frac{1}{2}kx^2=mgh$$

substitute in:

$$\frac{1}{2}(100)x^2=(2)(9.8)(4sin(30))$$

$x = 0.88543$ m

B) This one is like the previous one but you have to factor in the work done by friction. The two sides of the equation represent work in and work out (left side is out, the right side is in).

$$\frac{1}{2}kx^2+Fd = mgh$$

expand it once

$$\frac{1}{2}kx^2+uNd = mgh$$

substitute in

$$\frac{1}{2}(100)x^2+(0.2)(2)(9.8)(cos(30))(4)=(2)(9.8)(4sin(30))$$

$x = 0.68$ m

C) The work done in compressing the spring to that 68 centimeters is

$$\frac{1}{2}(100)(0.68)=34 Joules$$

.

Once the spring is "released" that energy will go into kinetic energy until all the kinetic energy has either been converted to potential energy (the height of the blocks) or has been canceled by the work done on the block by friction. If the block moves a distance x up the plane, its height increases by x

$$sin(30)=(\frac{1}{2})x$$

so its potential energy increases by

$$(2)(9.81)(\frac{x}{2})=0.91x$$

and the work done by friction is 3.4x. We must have 9.81x+ 3.4x = 13.21x= 34.

Well, that's easy: the block will move

$$\frac{34}{13.21}=2.57m$$

which is 2.57 - 0.68= 1.89 m. above the equilibrium point for the spring and 4 - 1.89= 3.11 meters below the original position of the block.