6) In this new scenario, what would the coefficient of friction of the rough pat

Question

6)

In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

7)

Return to a scenario where the blcok makes it throgh the entire rough patch. If the rough patch is lengthened to a distance of three times longer, as the block slides through the entire distance of the rough patch, the magnitude of the work done by the force of friction is:

the same

three times greater

three times less

nine times greater

nine times less

Explanation / Answer

1) W=0.5*m*v^2 = 0.5*12*(3.9)^2 = 91.26

2)
v = 3.9m/s

SE = 1/2*k*x^2
KE = 1/2*m*v^2

Total Energy = SE = KE

1/2*k*x^2 = 1/2*m*v^2
k*x^2 = m*v^2
x^2 = m/k*v^2
x = v*sqrt(m/k)
= 3.9*sqrt(12/4678)
=0.197 m

3)
KE1 = KE2 + frictionWork
FW = KE1 - KE2
= 75.6 J

4)
F = uk*N = uk*m*g
W = F*d = (uk*m*g) * d
d = W/F = W/(uk*m*g)
= 75.6/(0.5*12*9.8)
= 1.28 m

5)
d = 0.64m

W = uk*m*g*d
W = 0.5*12*9.8*0.64
= 37.632 J

SE = W
1/2*k*x^2 = W
x = sqrt(2*W/k)
= sqrt(2*37.632/4678)
= 0.1268 m

6)
W = d*uk*m*g

let d = half of rough patch


change uk so that mass barely travels through rough patch instead of halfway
W =d*uk*m*g = (2d)*(x*uk)*m*g
x = 1/2
so reduce uk by half
new uk = 0.25

7)

three times greater

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