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Two objects are in motion, one has 1/2 the velocity of the other but twice the mass. External forces bring each mass to rest. Choose the correct statement.
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Correct Answer: A. Twice as much work is done on the smaller mass.
PHY 5.3
Simple way:
Use the Work-Energy Theorem where the variation of the kinetic energy of a system is equal to the work of the net force applied to the system. Thus the work is just the change in kinetic energy (i.e. final MINUS initial). But the final is 0 because the final velocity is 0 (at “rest”). So we get:
Work = -m(v2)/2
Looking at the equation, it is clear that doubling the velocity has a greater impact on the work as compared to doubling the mass.
Side Note: Friction always does negative work because it is opposite to the direction of motion. Similarly, work must be negative in this problem.
Alternate method:
Work = Force times distance (actually, displacement) = m (a) x
To determine the acceleration a (actually, the deceleration):
V2 = Vo2 + 2ax = 0 (each object has a final vel. of 0)
a = – (Vo2)/2x
Work = m [- (Vo2)/2x] x = -m(Vo2)/2
Now by testing the numbers given, you can see that the original velocity has a greater impact than the mass.
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