A block is pressed against a spring at the bottom of an incline and realsed. The

Question

A block is pressed against a spring at the bottom of an incline and realsed. The mass of the block is 0.2kg, the force constant of the spring is 600N/m, the spring is compressed 0.05m, and the angle of the incline is 30 degrees. (a) If there is not friction, how far up the inlcline does the block travel?

I correctly solved this the following way:

1/2kx^2=mgh

h=dsin(30)

1/2kx^2=mgdsin(30)

d=(1/2kx^2)/(mgsin(3)= .765m

Now suppose there is friction and the block goes 3/4 the distance you calculed in (a). What is the value of the frictional force?

There answer my professor gives is 0.33N. How did he calculate this?

Explanation / Answer

a) initial potential energy of spring = final gravitational potential energy of block kx^2 / 2 = mgy kx^2 / 2 = mgLsin? L = kx^2 / [2mgsin?] L = (1400 N/m)(0.100 m)^2 / [2(0.200 kg)(9.81 m/s^2)(sin60.0°)] L = 4.12 m b) initial potential energy of spring = (work done by friction) + (final gravitational potential energy of block) kx^2 / 2 = µNL + mgy kx^2 / 2 = µmgLcos? + mgLsin? L = kx^2 / [2mg(µ*cos? + sin?)] L = (1400 N/m)(0.100 m)^2 / [2(0.200 kg)(9.81 m/s^2)((0.400)*cos60.0° + sin60.0°)] L = 3.35 m

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