Question

A 1.70-m tall person lifts a 2.50-kg book from the ground so it is 2.40 m above the ground.Part AWhat is the potential energy of the book relative to the ground?Express your answer to three significant figures and include the appropriate units.Part BWhat is the potential energy of the book relative to the top of the person's head?Express your answer to three significant figures and include the appropriate units.Part CHow is the work done by the person related to the answers in parts A and B?1. The work done by the person in lifting the book from the ground to the final height is the same as the answer to part A2. The work done by the person in lifting the book from the ground to the final height is the same as the answer to part B

Answer 1

Answer:

Part a)

$U = 58.9 J$

Part b)

$U = 17.2 J$

Part C)

work done by the person to bring the book is same as the answer solved in part A)

Explanation:

Part a)

potential energy of book with respect to the ground position is given as

$U = mgh$

here we know that

m = mass of book

g = 9.81 m/s/s

h = height of the book with respect to the initial reference level

$U = (2.50)(9.81)(2.40)$

$U = 58.9 J$

Part b)

now we have to find the potential energy of the book with respect to the top head position of the person

$U = mgh'$

here we know

$h' = h - h_1$

$h' = 2.40 - 1.70$

$h' = 0.70 m$

now from above formula we have

$U = (2.50)(9.81)(0.70)$

$U = 17.2 J$

Part C)

work done by the person to put the book at given height is given as

$W = F.d$

here we know that

$F = mg$

$d = 2.40 m$

$F = (2.50)(9.81) = 24.5 N$

Now work done is given as

$W = 24.5(2.40) = 58.9 J$

so work done by the person to bring the book is same as the answer solved in part A)