All Solutions
Page 265: Section Review
$$
W=Fs
$$
$$
W=80cdot10
$$
$$
boxed{W=800,,rm J}
$$
W=800,,rm J
$$
$$W=Fs$$
in this case force is equal to weight of the refrigerator and distance is equal to height.
$$W=mghcdot sin (theta )$$
$$W=185cdot9.81cdot10 cdot sin (11,,^{o} )$$
$$boxed{W=3460,,rm J}$$
$$
W=Fs
$$
We can see that work only depends on force and distance and if we change how fast we move the book it will not change the work done on the book, answer on this question is $textbf{no}$ it does not depend on how fast you rise it.
Power is defined as work per time:
$$
P=frac{W}{t}
$$
Now we can see that power depends on time and if we move the book faster time will decrease, answer on this question is $textbf{yes}$ it does depend on how fast you rise it.
$$W=Fs$$
in this case force is equal to weight and distance is equal to height.
And now when we have work we can get power:
$$P=frac{W}{t}$$
When we connect those equations we get:
$$P=frac{mgh}{t}$$
$$P=frac{1.1cdot10^{3}cdot9.81cdot 40}{12.5}$$
$$boxed{P=34530,,rm W}$$
$$
W=Fs
$$
in this case force is equal to weight and distance is equal to height.
$$
W=mgh
$$
$$
W=0.18cdot9.81cdot2.5
$$
$$
boxed{W=4.41,,rm J}
$$
$$
W=Fs
$$
In this case force is equal to weight and distance is equal to height.
$$
W=mghrightarrow m=frac{W}{gh}
$$
$$
m=frac{7cdot10^{3}}{9.81cdot1.2}
$$
$$
boxed{m=595,,rm kg}
$$
from the first floor of a building to a room located
on the second floor, farther down the hall. You
choose to carry the box first up the stairs, and then
down the hall to the room. Your friend carries it
down the hall on the first floor, then up a different
stairwell to the second floor. Question is who does more work.
$$
W=vec Fcdotvec s
$$
In this case to lift a box we have to overcome gravitational force. That means if we go down our force on the box will be in opposite direction that we are moving and work will be negative, if we go up our force will be in the same direction and work will be positive, this means that it does not matter if we move in any direction as long we and up in the same spot total work will be the same. Answer to this question is you and your friend will do the same work on the box.
We can write relation between velocity and kinetic energy with equation:
$$
E_k=frac{mv^2}{2}rightarrow v=sqrt{frac{2E_k}{m}}
$$
now we can see that if kinetic energy is doubled, velocity is not. When we put that $E_{k2}=2E_k$ we get:
$$
v_2=sqrt{frac{2E_{k2}}{m}}=sqrt{frac{2cdot2E_{k}}{m}}=sqrt{2}sqrt{frac{2E_{k}}{m}}
$$
$$
boxed{v_2=sqrt{2}v}
$$
v_2=sqrt{2}v
$$
