### Problems & Exercises

(a) $5\text{.}\text{92}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}$

(b) $-5\text{.}\text{88}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}$

(c) The net force is zero.

(a) $1\text{.}\text{96}\times {\text{10}}^{\text{16}}\phantom{\rule{0.25em}{0ex}}\text{J}$

(b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb.

Equating ${\text{\Delta PE}}_{\mathrm{g}}$ and $\text{\Delta KE}$, we obtain $v=\sqrt{2\text{gh}+{{v}_{0}}^{2}}=\sqrt{2(\text{9.80 m}{\text{/s}}^{2})(\text{20.0 m})+(\text{15.0 m/s}{)}^{2}}=\text{24.8 m/s}$

(a) $\text{25}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{years}$

(b) This is much, much longer than human time scales.

(a) $9\text{.}\text{46}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{J}$

(b) $2\text{.}\text{54 y}$

Identify knowns: $m=\text{950 kg}$, $\text{slope angle}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{2.00\xba}$, $v=\mathrm{3.00\; m/s}$, $f=\text{600 N}$

Identify unknowns: power $P$ of the car, force $F$ that car applies to road

Solve for unknown:

$P=\frac{W}{t}=\frac{\text{Fd}}{t}=F\left(\frac{d}{t}\right)=\text{Fv},$

where $F$ is parallel to the incline and must oppose the resistive forces and the force of gravity:

$F=f+w=\text{600 N}+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $

Insert this into the expression for power and solve:

$\begin{array}{lll}P& =& \left(f+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)v\\ & =& \left[\text{600 N}+\left(\text{950 kg}\right)\left({\text{9.80 m/s}}^{2}\right)\text{sin 2\xba}\right](\text{30.0 m/s})\\ & =& \text{2.77}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\mathrm{W}\end{array}$

About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline.

(a) $3\text{.}\text{21}\times {\text{10}}^{4}\text{N}$

(b) $2\text{.}\text{35}\times {\text{10}}^{3}\text{N}$

(c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b)

(a) $2\text{.}\text{50}\times {\text{10}}^{\text{12}}\text{J}$

(b) 2.52%

(c) $1\text{.}4\times {\text{10}}^{4}\text{kg}$ (14 metric tons)

(a) $\text{0.500 m}{\text{/s}}^{2}$

(b) $\mathrm{62.5\; N}$

(c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since $f=F-\text{ma}$. If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared (${t}^{2}$). Therefore, the water resistance will not depend linearly on the velocity.

(a) $\text{16}\text{.}1\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\mathrm{N}$

(b) $3\text{.}\text{22}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\mathrm{J}$

(c) 5.66 m/s

(d) 4.00 kJ

(a) $4\text{.}\text{65}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kcal}$

(b) 38.8 kcal/min

(c) This power output is higher than the highest value on Table 7.5, which is about 35 kcal/min (corresponding to 2415 watts) for sprinting.

(d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!).

(a) 4.32 m/s

(b) $3\text{.}\text{47}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\mathrm{N}$

(c) 8.93 kW

(a) Box A has the greatest speed since it has the greatest work done; work equals change in kinetic energy and thus fastest speed.

(b) i. Yes, this equation is consistent. An increased force or distance increases the work done, and thus increases the change in kinesthetic energy. ii. No, this equation does not make sense.

(c)

### Test Prep for AP® Courses

The kinetic energy should change in the form of –cos, with an initial value of 0 or slightly above, and ending at the same level.

Any force acting perpendicular will have no effect on kinetic energy. Obvious examples are gravity and the normal force, but others include wind directly from the side and rain or other precipitation falling straight down.

Note that the wind is pushing from behind and one side, so your KE will increase. The net force has components of 1400 N in the direction of travel and 212 N perpendicular to the direction of travel. So the net force is 1420 N at 8.5 degrees from the direction of travel.

Gravity has a component perpendicular to the cannon (and to displacement, so it is irrelevant) and has a component parallel to the cannon. The latter is equal to 9.8 N. Thus the net force in the direction of the displacement is 45 N − 9.8 N, and the kinetic energy is 53 J.

The potato cannon (and many other projectile launchers) above is an option, with a force launching the projectile, friction, potentially gravity depending on the direction it is pointed, etc. A drag (or other) car accelerating is another possibility.

The kinetic energy of the rear wagon increases. The front wagon does not, until the rear wagon collides with it. The total system may be treated by its center of mass, halfway between the wagons, and its energy increases by the same amount as the sum of the two individual wagons.