Can someone help me set up this equation? Thank you will award points ASAP A 3 k
ID: 1285615 • Letter: C
Question
Can someone help me set up this equation? Thank you will award points ASAP
A 3 kg block travels along a horizontal surface with a coefficient of kinetic friction of 0.18 at a speed of 7 m/s. After sliding a distance of 1.8 m the block makes a smooth transition to a ramp with a coefficient of kinetic friction of 0.18. How far up the ramp does the block travel before coming to a momentary stop.
Can someone help me set up this equation? Thank you will award points ASAP A 3 kg block travels along a horizontal surface with a coefficient of kinetic friction of 0.18 at a speed of 7 m/s. After sliding a distance of 1.8 m the block makes a smooth transition to a ramp with a coefficient of kinetic friction of 0.18. How far up the ramp does the block travel before coming to a momentary stop.Explanation / Answer
acceleration on flat surface,
a = -g*mue_k
= -9.8*0.18
= -1.764 m/s^2
let v is the speed at the bottom of the ramp.
v^2 - u^2 = 2*a*s
v = sqrt(u^2 + 2*a*s)
= sqrt(7^2 -2*1.764*1.8)
= 6.54 m/s
acceleration on inclined surface,
a = -g*sin(40) - mue_k*g*cos(40)
= -9.8*sin(40) - 0.18*9.8*cos(40)
= -7.65 m/s^2
final speed, v' = 0
now use, v'^2 - v^2 = 2*a*d
d = (v'^2-v^2)/(2*a)
= (0^2 - 6.54^2)/(2*(-7.65))
= 2.795 m <<<<<<---------Aswer
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