Day: September 27, 2022

Posted on

202209271040 Exercise 8.4 (Q53)

Neglecting the Earth’s rotation, show that the energy needed to launch a satellite of mass m into circular orbit at altitude h is

\displaystyle{\bigg( \frac{GM_\textrm{E}m}{R_{\textrm{E}}}\bigg)\bigg(\frac{R_{\textrm{E}}+2h}{2(R_{\textrm{E}}+h)}\bigg)}.

Extracted from R. Wolfson. (2016). Essential University Physics.

Abortive attempt.

(energy/work-done approach)

By conservation of mechanical energy,

\begin{aligned} \Delta (\textrm{KE}+\textrm{PE}) & = 0 \\ (\textrm{KE}_f-\textrm{KE}_i) + (\textrm{PE}_f-\textrm{PE}_i) & = 0 \\ \bigg(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\bigg) + \big(mg_f(R_\textrm{E}+h)-mg_iR_\textrm{E}\big) & = 0 \\ \end{aligned}

as

\begin{aligned} g_f & = G\frac{M_\textrm{E}}{(R_\textrm{E}+h)^2} \\ g_i & = G\frac{M_\textrm{E}}{(R_\textrm{E})^2} \\ \end{aligned}

On the one hand, the gravitational pull provides the centripetal force for revolving at the orbital speed v:

\begin{aligned} \text{}_MF_m & = \text{}_mF_M \\ \frac{mv^2}{R_{\textrm{E}}+h} & = G\frac{mM_\textrm{E}}{(R_{\textrm{E}}+h)^2} \\ v & = \sqrt{\frac{GM_\textrm{E}}{R_{\textrm{E}}+h}} \\ \end{aligned}

On the other hand, the escape speed u of the satellite is

\begin{aligned} \frac{1}{2}mu^2 & = G\frac{mM_\textrm{E}}{R_\textrm{E}} \\ u & = \sqrt{\frac{2GM_\textrm{E}}{R_\textrm{E}}} \\ \end{aligned}

but what is this question asking for?

\Delta\textrm{KE}=\displaystyle{\frac{m(v^2-u^2)}{2}}

(to be continued)

Posted on

202209271004 Exercise 4.5 (Q31)

An elevator accelerates downward at 2.4\,\mathrm{m\, s^{-2}}. What force does the elevator’s floor exert on a 52\,\mathrm{kg} passenger?

Extracted from R. Wolfson. (2016). Essential University Physics.

Roughwork.

Take downward positive, \downarrow\textrm{+ve}. By Newton’s 2nd Law,

\mathbf{F}_{\textrm{net}}=m\mathbf{a}.

Given m=52\,\mathrm{kg}. Let F be the unknown magnitude of force:

\begin{aligned} \mathbf{a} & = 2.4\,\hat{\mathbf{k}} \\ \mathbf{F}_\textrm{net} & = -F\,\hat{\mathbf{k}}+mg\,\hat{\mathbf{k}} \\ & = (-F+mg)\,\hat{\mathbf{k}} \\ \end{aligned}

Then,

\begin{aligned} -F+mg & = 2.4m \\ F & = (g-2.4)m \\ & = (9.81-2.4)(52) \\ & = 385\,\mathrm{N}\qquad \textrm{(3 s.f.)} \end{aligned}

\therefore The force the elevator’s floor exerts on the passenger is 385\,\mathrm{N} upward.

Posted on

202209270930 Exercise 5.3.B (Q1-Q6)

1. What is the potential energy of a 10\,\mathrm{kg} mass 25 metres above the ground?
2. A 3\,\mathrm{kg} mass is 20\,\mathrm{m} above the ground. How much potential energy does it have?
3. How high must you raise a 25\,\mathrm{kg} mass before it has a potential energy of 8500\,\mathrm{J}?
4. How high must you raise a 2\,\mathrm{kg} mass before it has a potential energy of 10\,\mathrm{J}?
5. A mass has a potential energy of 1000\,\mathrm{J} when it is 40\,\mathrm{m} above ground level. What is the mass?
6. A mass has a potential energy of 20\,\mathrm{J} when it is 50\,\mathrm{cm} above ground level. What is the mass?

Extracted from B. Kennedy. (1999). Progressive Problems for ‘S’ Grade Physics.

Background.

\textrm{PE}=mgh

Roughwork.

1.

\begin{aligned} \textrm{PE} & = mgh \\ & = (10)(9.81)(25) \\ & = 2450\,\mathrm{J}\qquad \textrm{(3 s.f.)} \\ \end{aligned}

2.

\begin{aligned} \textrm{PE} & = mgh \\ & = (3)(9.81)(20) \\ & = 589\,\mathrm{J}\qquad \textrm{(3 s.f.)} \end{aligned}

3.

\begin{aligned} \textrm{PE} & = mgh \\ 8500 & = (25)(9.81)(h) \\ h & = 34.7\,\mathrm{m}\qquad \textrm{(3 s.f.)} \end{aligned}

4.

\begin{aligned} \textrm{PE} & = mgh \\ 10 & = (2)(9.81)(h) \\ h & = 51.0\,\mathrm{cm}\qquad \textrm{(3 s.f.)} \end{aligned}

5.

\begin{aligned} \textrm{PE} & = mgh \\ 1000 & = (m)(9.81)(40) \\ m & = 2.55\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}

6.

\begin{aligned} \textrm{PE} & = mgh \\ 20 & = (m)(9.81)(0.5) \\ m & = 4.08\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}