4. An object of mass 0,550 kg is lifted from the floor to a height of 3.50 m at

Question

4. An object of mass 0,550 kg is lifted from the floor to a height of 3.50 m at constant speed. How much work is done by the lifting force (include units)? How much work is done by the gravitational force on the object? What is the net work done on the object? What is the change in kinetic energy of the object? Are your results consistent with the work-energy principle? Explain. If the object in Question 4 is released from rest after it is lifted, what is its ki- netic energy just before it hits the floor? What is its velocity? Show your work and include units. Answers: 5. Kinetic energy: Velocity: A force acts on an object of mass 0.425 kg. The force varies with position as shown in the graph that follows. 6. REALTIME PHYSICS: MECHANICS 250 WILEY www.wiley.com/college/sokoloff

Explanation / Answer

4) Work done by the lifting force W = F * d = m g d

= 0.55 * 3.5 * 9.8 = 18.87 J

Net work done on the object = 0 J

Change in kinetic energy of the object

As velocity is constant

W = delta E = 0

Yes results are consistent with work energy theorem

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5) KE = m g h = 0.55 * 3.5 * 9.8 = 18.87 J

KE = 1/2 m v2

18.87 J = 1/2 * 0.55 v2

v = 8.3 m/s

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