AP Physics 1- 3.5 Power - Exam Style questions - FRQs- New Syllabus

(a)
The surface gravitational field is proportional to \( \dfrac{M}{R^2} \).

Therefore,

\( g_{\text{Mars}} = \dfrac{0.11}{(0.53)^2}g \)

\( g_{\text{Mars}} \approx \dfrac{0.11}{0.2809}g \approx 0.392g \)

So,

\( \boxed{g_{\text{Mars}} \approx 0.39g} \)

(b)
Sojourner’s weight on Mars is

\( W = mg_{\text{Mars}} \)

Using \( g_{\text{Mars}} \approx 0.39(9.8) = 3.82\ \mathrm{m/s^2} \),

\( W = (11.5)(3.82) \approx 43.9\ \mathrm{N} \)

Therefore,

\( \boxed{W \approx 44\ \mathrm{N}} \)

(c)
On a ramp inclined at \( 20^\circ \), the normal force is

\( F_N = W\cos\theta \)

\( F_N = (44)\cos 20^\circ \)

\( F_N \approx 44(0.94) \approx 41.3\ \mathrm{N} \)

So the minimum normal force required is

\( \boxed{41\ \mathrm{N}} \)

The ramp only needs to balance the component of the weight perpendicular to the surface.

(d)
The net force is

\( \boxed{0\ \mathrm{N}} \)

because the rover moves at constant velocity, which means its acceleration is zero. By Newton’s second law, zero acceleration means zero net force.

(e)
The stored energy is used at a rate of \( 10\ \mathrm{W} = 10\ \mathrm{J/s} \).

So the total operating time is

\( t = \dfrac{E}{P} = \dfrac{5.4\times 10^5}{10} = 5.4\times 10^4\ \mathrm{s} \)

The distance traveled is

\( d = vt \)

\( d = (6.7\times 10^{-3})(5.4\times 10^4) \)

\( d = 361.8\ \mathrm{m} \)

Therefore,

\( \boxed{d \approx 3.6\times 10^2\ \mathrm{m}} \)

or about \( \boxed{362\ \mathrm{m}} \).

(f)
\( 0.010\% = 0.00010 \) as a decimal fraction.

So the power used against drag is

\( P_{\text{drag}} = 0.00010(10) = 1.0\times 10^{-3}\ \mathrm{W} \)

Use \( P = Fv \):

\( F_{\text{drag}} = \dfrac{P_{\text{drag}}}{v} = \dfrac{1.0\times 10^{-3}}{6.7\times 10^{-3}} \)

\( F_{\text{drag}} \approx 0.149\ \mathrm{N} \)

Therefore,

\( \boxed{F_{\text{drag}} \approx 0.15\ \mathrm{N}} \)