a.) How fast is the block when it reaches the bottom of the ramp? b.) What is th
ID: 1780369 • Letter: A
Question
a.) How fast is the block when it reaches the bottom of the ramp?
b.) What is the maximum distance that the spring is compressed by the box?
c.) Would the answer to b.) be different, if instead of a 0.5 meter section on the flat ground with k = 0.4, there was instead a 0.5 meter section on the inclined ramp with k = 0.4?
2) (6.5 points) A 2.5 kg box is released from rest 1.5 meters above the ground and slides down a frictionless ramp. It slides across a floor that is frictionless, except for a small section 0.5 meters wide that has a coefficient of kinetic friction of 0.4. At the left end, is a spring with spring constant 250 N/m. The box compresses the spring, and then is accelerated back to the right. 2.5 k k=250 N/m h=1.5 m He = 0.4 d=0.50 mExplanation / Answer
2)
initial mechanical energy Ei = m*g*h
final mechanical energy at the bottom Ef = (1/2)*m*v^2
from energy conservation
ttoal energy is conserved
Ef = Ei
(1/2)*m*v^2 = m*g*h
v = sqrt(2*g*h)
speed v = sqrt(2*9.8*1.5) = 5.42 m/s
====================
work done by friction Wf = -uk*m*g*d
initial mechanical energy Ei = m*g*h
final mechanical energy Ef = (1/2)*k*x^2
wor energy theorem
Work = Ef - Ei
-uk*m*g*d = (1/2)*k*x^2 - m*g*h
-0.4*2.5*9.8*0.5 = (1/2)*250*x^2 - (2.5*9.8*1.5)
x = 0.505 m <<<<<-----ANSWER
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work done by friction
Wf = -uk*m*g*d*costheta
-uk*m*g*d*costheta = (1/2)*k*x1^2 - m*g*h
-0.4*2.5*9.8*0.5*costheta = (1/2)*250*x1^2 - (2.5*9.8*1.5)
(1/2)*250*x1^2 = (2.5*9.8*1.5) - 0.4*2.5*9.8*0.5*costheta
x1 > x
the compression is more in c than in b
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