Use the worked example above to help you solve this problem. A 15, 000 N car sta
ID: 1607228 • Letter: U
Question
Use the worked example above to help you solve this problem. A 15, 000 N car starts from rest and rolls down a hill from a height of 10.0 m see figure It then moves across a level surface and collides with a light spring-loaded guardrail. (a) Neglecting any losses due to friction, and ignoring the rotational kinetic energy of the wheels, find the maximum distance the spring is compressed. Assume a spring constant of 1.5 times 10^6 N/m. (b) Calculate the maximum acceleration of the car after contact with the spring, assuming no frictional losses. (c) If the spring is compressed by only 0.30 m, find the change in the mechanical energy due to friction. A spring-loaded gun fires a 0.090-kg puck along a tabletop. The puck slides up a curved ramp and flies straight up into the air. (a) If the spring is displaced 22.0 cm from equilibrium and the spring constant is 875 N/m, how high does the puck rise, neglecting friction? (b) If instead it only rises to a height of 5.00 m because of friction, what is the change in mechanical energy?Explanation / Answer
from conservation of energy
initial energy = final energy
m*g*h = (1/2)*k*x^2
15000*10 = (1/2)*1.5*10^6*x^2
x = 0.447 m
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(b)
maximum acceleration amax = w^2*x =( k/m)*x
mass m = 15000/9.8 kg
amax = (1.5*10^6*9.8/15000)*0.447 = 438.1 m/s^2
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(c)
change in mechanical energy = m*g*h - (1/2)*k*x1^2
change in mechanical energy Wnc= (15000*10) - ((1/2)*1.5*10^6*0.3^2) = 82500 J
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(a)
initial elastic potential energy = final gravitational potential energy
(1/2)*k*x^2 = m*g*h
(1/2)*875*0.22^2 = 0.09*9.8*h
h = 24 m
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(b)
change in mechanical energy = (1/2)*k*x^2 - m*g*h1
change in mechanical energy Wnc= (1/2)*875*0.22^2 - (0.09*9.8*5) = 16.765 J
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