A 12.5-kg crate slides along a horizontal frictionless surface at a constant spe
ID: 1395782 • Letter: A
Question
A 12.5-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure.
a) What is the kinetic energy of the crate as it slides on the upper surface?
b) While the crate slides along the upper surface, how much gravitational potential energy does it have compared to what it would have on the lower surface?
c) What is the speed of the crate when it arrives at the lower surface?
d) What is the kinetic energy of the crate as it slides on the lower surface?
e) What minimum coefficient of kinetic friction is required to bring the crate to a stop over a distance of 5.0 m along the lower surface?
Explanation / Answer
a)
KE at the upper surface
KE1 = 0.5*m*v1^2 = 0.5*12.5*4^2 = 100 J
b)
PE = m*g*h = 12.5*9.8*3 = 367.5 J
c)
kinetic energy at the botom = KE2 = 0.5*m*v2^2
PE2 = 0
total energy at the upper surface TE1 = KE1 + PE1
total energy at the lower surface TE2 = KE2 + PE2 = KE2
fro energy conservation
TE1 = TE2
0.5*m*v2^2 = 100+367.5
v2 = 8.65 m/s <----answer
d)
KE2 = 0.5*m*v2^2 = 467.5 J
e)
normal force on the plane = N = m8g*costheta = m*g*(4/5) =
12.5*9.8*(4/5) = 98 N
frictional force f= uk*N
work done by rictional force = wf = fk*s = uk*98*5 = uk*490
KE2 = 0
from work enrgy theorem
TE1 - Wf = 0
TE1 = Wf
uk = 467.5/490 = 0.95
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