**SOLUTIONS FOR HOMEWORK #7**
1. W = F x d

If the applied force is 20N and the distance through which the force is applied is 3.5m, then:

W = F x d = 20 N x 3.5 m = 70 Nm = 70 J
2.

a) W = F x d = 90 N x 3 m = 270 J; the block has also acquired the same amount of potential energy, since its potential energy
is
its weight times its height above the ground.

b) This is an example of an inclined plane acting as a machine. At this angle, you need apply only 54 N of force, but must travel
through 5 meters. Still, W = F x d = 54 N x 5 m = 270 J, as above. The mass is still only 3 m above the ground, so its PE is still
only 270J, in other words, its gain in PE equals the amount of work done on it.

3.

a) KE = 1/2 mv^{2} = 1/2 (3 kg) (4 m/s)^{2}=

1/2 (3 kg) (16 m^{2}/s^{2} = 24J

b) Here, you could substitute 8 m/s for v and find that the new KE is 96 J, or you could reason as follows: I know that KE is
proportional to the square of v. So, if v doubles, then v^{2} increases by a factor of 2^{2} or a factor of four, so the
new kinetic energy must be four times greater than the original kinetic energy. Since the KE in part a) was 24J, the KE now is 4 times
that or 4x24J = 96J. (see the next to last problem for a slightly different variant of this).

4.

a) Here you are told that for every 2m of rope you pull, the piano rises only 0.4 m, or in other words, you have to pull five times as
much rope as you lift the piano. This means you have to exert your force through five times as much distance as you raise the piano.
This means that you have to exert only 1/5 as much force as the weight of the piano.

Force applied to piano x distance through which this force is applied = weight of piano x height piano rises

Divide terms to create this ratio:

(Distance through which force is applied/height piano rises)= (weight of piano/force applied to piano)

If the left hand side is equal to 5, then the piano weighs five times more than the force you need to apply. Since we are given that the
piano is 5000N, you need apply only 1000N.

b) In part a) we calculated the force needed with no friction. We are told that with friction, we actually use 2500N. The efficiency
of this machine is then 1000N/2500N = 40%.
5. Extra strectching means that there is more tension in the rubber band and therefore more b> force on the rock. Additionally,
more stretching means the force is exerted through a greater ** distance **. Remembering that Work = F x d, increasing F and d
increases the work done on the rock, which increases its elastic PE, which means it will have greater KE and greater speed when the rock
is released.

6. We are told that the mouse and elephant have the same KE. Remember that KE = 1/2 mv^{2}, we can then write:

KE_{mouse} = 1/2 m_{mouse } v_{mouse}^{2}

KE_{elephant} = 1/2 m_{elephant} v_{elephant}^{2}

Equating these two expressions and dividing through by 1/2 gives us:

m_{elephant} v_{elephant}^{2} = m_{mouse } v_{mouse}^{2}

Since we know the mass of the elephant is hugely greater than the mass of the mouse, the only way this product can be equal is if the
velocity of the mouse is much larger than the velocity of the elephant.
7. The key to this question is the conservation of energy. The satellite will have at each point in its orbit a different speed, so at
each point it will have a different kinetic energy. Also, you are told that PE increases as you get farther from the center of the
Earth (as you already know). So, the total energy of the satellite at each point in its orbit is:

Total energy = KE + PE

When the PE increases, KE must decrease; when the PE decreases, KE increases. The *minimum* PE occurs when the
satellite is nearest the Earth. This means that the * maximum * KE occurs when the satellite is nearest the Earth. Since
KE is proportional to v^{2}, the greatest speed occurs when the KE is greatest, which is at closest approach to the Earth.
(In astronomy, the point of an orbit closest to the Earth is known as **perigee**).
8. PE = mgh = 1000 N x 5 m = 5000 J (remember, Newtons measure weight, so the 1000 N equals mg).

The PE at the top equals the KE at the bottom. KE = 1/2 mv^{2}

We can equate PE at the top to KE at the bottom to get:

mgh = 1/2mv^{2}

Dividing through by m and rearranging, we have:

v = sqrt (2gh)

where sqrt means square root.

Solving:

v = sqrt (2 x 10 m/s/s x 5 m) = sqrt(100 mxm/sxs) = 10 m/s
9. If the hammer were taken to a height that is four times greater, it would have four times more potential energy. If it were
released from that point, it would strike the ground with four times more kinetic energy. Remembering that KE is proportional to
v^{2}, the speed doubles if the KE quadruples (see the third problem on this assignment).

10. The mechanical advantage is two, meaning the amount of force you need to apply is only half the weight of the object, since there
are two strings support the weight in this pulley arrangement.