ll. Lifting a book In this problem, you\'ll use the formal definition of workto
ID: 1393517 • Letter: L
Question
ll. Lifting a book In this problem, you'll use the formal definition of workto figure out some stuff Workis done when a force Facts on the object over a distance Ax. When the force pointsin the same direction that the object moves, the work is given by W FAc. (In lecture, you'll deal with misaligned" forces and isplacements.) A. A student holds a book ofmass m in her hand and raises the book vertically at constant speed. Sketch a free-body diagram for the book. As the book rises at steady speed, is the force exerted by the student on the book greater than, less than, or equal to mg? Explain briefly B. Suppose the student does 25 joules of work lifting the book. 1. Does the book lifted at constant speed gain potential energy, kinetic energy, or both? Explain. 2. Is the potential energy gained by the book greater than, less than, or e to 25 joules? Explain. qual C. Now we'll repeat the reasoning of part Binterms of symbols rather than numbers 1. Use the definition of work to determine the amount of work the student does in raising the book through a height h. Express your answerin terms of m, g, and h 2. So, how much potential energy did the book gain, in tenms of m, g, and h? D. In this class or a previous class, you mayhave seen the equation U mgh for gravitational potential energy. For people who alreadyknew that formula, what's the point of parts Band Cabove?Explanation / Answer
A) As the book is mvoing with constant speed, Net force = 0
F_applied - m*g = 0
F_applied = m*g
B)
1)Only potentail enrgy.Because, there is no change in kinetic enrgy
2)equal to 25 J. on Applying conservation of Enrgy we get gain in PE = 24 J.
C)
1) Workdone by the student = F_applied*h
= m*g*h
2) gain in PE = m*g*h
d) Whe the book is moved with constant speed then only these values are correct.
otherwise the workdone by person on the book will not be equal to m*g*h.
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