A 1.7kg book is lying on a 0.80m -high table. You pick it up and place it on a bookshelf 2.3m above the floor.
A) During this process, how much work does gravity do on the book?
B) During this process, how much work does your hand do on the book?
Answer
(A) The expression to calculate the work done is, W=F \cdot d \cos \theta
When the book is picked up and placed on a book shelf, the distance is equal to the difference of the heights.
\begin{aligned}
d=2.3 \mathrm{~m}-0.80 \mathrm{~m} \\
=1.50 \mathrm{~m}
\end{aligned}The force is calculated as follows:
F=m g
Here, \mathrm{m} is the mass, \mathrm{g} is the gravity, and \mathrm{F}0 is the weight force.
Substitute 1.7 \mathrm{~kg} for \mathrm{m} and 9.8 \mathrm{~m} / \mathrm{s}^{2} for \mathrm{g} in expression F=m g.
\begin{aligned}
F=(1.7 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \\
=16.67 \mathrm{~N}
\end{aligned}Substitute 16.66 \mathrm{~N} for \mathrm{F}, 1.50 \mathrm{~m} for \mathrm{d}, and 180^{\circ} for \theta in expression W=F \cdot d \cos \theta \mathrm{v}
\begin{aligned}
W=(16.67 \mathrm{~N})(1.50 \mathrm{~m}) \cos 180^{\circ} \\
=-25.0 \mathrm{~J}
\end{aligned}Part A The work done by the gravity on the book is -25.0 \mathrm{~J}.
(B) The force is calculated as follows:
F=m g
Here, m is the mass, g is the gravity, and F is the weight force.
Substitute 1.7 \mathrm{~kg} for \mathrm{m} and -9.8 \mathrm{~m} / \mathrm{s}^{2} for g in expression F=m g.
\begin{aligned}
F=(1.7 \mathrm{~kg})\left(-9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \\
=-16.67 \mathrm{~N}
\end{aligned}Substitute -16.67 \mathrm{~N} for \mathrm{F}, 1.50 \mathrm{~m} for \mathrm{d}, and 180^{\circ} for \theta in expression W=\vec{F} \cdot d \cos \theta.
\begin{aligned}
W=(-16.67 \mathrm{~N})(1.50 \mathrm{~m}) \cos 180^{\circ} \\
=25.0 \mathrm{~J}
\end{aligned}Part B The work done by the hand on the book is 25.0 \mathrm{~J}.
