A block starts from rest at a height of 8.7m on a fixed inclined plane. The spee

Question

A block starts from rest at a height of 8.7m on a fixed inclined plane. The speed of the block at the bottom of the ramp is =take along the slope
driving force= mgsin22- 0.2mgcos22
=4.26N
length of slope=s=8.7/sin22=23.22
a=4.26/2.3=1.85
v^2= 2as
v=9.27
The acceleration of gravity is 9.8 m/s^2.
(U (mu) = 0.2 Weight of block=2.3kg Theta=22 degrees)
If the block continues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground until coming to rest? Answer in units of m

Explanation / Answer

a = 9.8*0.2 = 1.96 m/s^2 v^2 - u^2 = 2as 9.27^2 = 2*1.96*s s = 21.92 m

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