Question

As shown below (not to scale), a block of mass starts from rest and slides down a frictionless ramp of height h. Upon reaching the bottom of the ramp, it continues to slide across a flat frictionless surface. It then crosses a "rough patch" on the surface of length d=10m. This rough patch has a coefficient of kinetic friction uK=.1. After crossing the rough patch, the block's final speed is vf=2m/s. What is the height of the ramp? Hint: I would use the general conservation of energy expression where Wfriction is the work done by the force of friction and Ef and Ei are the total final and initial energies of the box respectively.

W friction=Ef-Ei

13. (10 points) As shown below and slides (not to scale), a block of mass m starts from rest down a frictionless ramp the it surface to of h. Upon reaching the bottom of ramp, the of slide across a flat frictionless "rough patch" on 0.1. After length surface. It then crosses a puk height of the d 10 m. This rough patch has a coefficient of kinetic friction the crossing the rough patch, the block's final speed is vf 2 m/s What is would use the general conservation of energy expression ramp? nt: I Wfric final where Weric work done by the force of friction and Er and Ei are the total is the and initial energies of the box respectively.

Answer 1

on rough patch,

fk = uk N = uk m g

a = - uk m g / m = - uk g = - 0.98 m/s^2

for rough patch, (d = 10 m)

vf^2 - vi^2 = 2 a d

2^2 - v^2 = 2 x - 0.98 x 10

v = 4.86 m/s

now applying energy conservation,

m g h = m v^2 /2

9.8 x h = 4.86^2 /2

h = 1.20 m