On an incline a 2.00 kg block is released 4.00m from a long spring with a force constant k= 60.0 N/m that is attached at the bottom of the incline. The incline makes the angle of 53 degrees with the horizontal, and the coefficients of friction between the block and the incline are u static = .40 and uk= .20.
a) what is the speed of the block just before it reaches the spring?
b.) what will be the maximum compression of the spring?
c.) the block rebounds back up the incline. How close does it get to its initial position?© BrainMass Inc. brainmass.com February 24, 2021, 2:14 pm ad1c9bdddf
work done by the block against the friction to move 4 m along the plane
Wf = uk*m*g*cos(53) = 0.2*2*9.8*0.6 = 2.352 J
work done by the gravity Wg = m*g*sin(53) = 2*9.8*0.8 = 15.68 J
Net gain of K.E. energy = Wg - Wf = 15.68 - 2.352 = 13.328 J
this gain of K.E. enegy will provide the velocity to ...
The solution shows all the formulas and calculations to arrive at the answers. The speed of a block, maximum compression of springs and blocks of rebounds are given.