Concept Of Physics The Forces

H C Verma

Concept Of Physics

1.   1. The gravitational force acting on a particle of $1$ g due to a similar particle is equal to $ 6.67$ x $10^{-17} N.$ Calculate the separation between the particles.

$m$ = $1 gm$ = $\frac{1}{1000}$kg $\\$ $F$ = $6.67$ x $10^{-17}$ $N$ $\Rightarrow$ $F$ = $\frac{Gm_1m_2}{r^2}$ $\\$ $\therefore$ $6.67$ x $20^{-17}$ = $\frac{6.67\times10^{-11} \times (\frac{1}{1000})\times (\frac{1}{1000})}{r^2}$ $\\$ $\Rightarrow$ $r^2$ = $\frac{6.67\times 10^{-11} \times 10^{-6}}{6.64\times10^{-17}}$ = $\frac{10^{-17}}{10^-{17}}$ = $1$$\\$ $\Rightarrow$ r =$ \sqrt1$ = 1 metre. $\\$ So. The separation between the particle is $1 m $.

2.   2. Calculate the force with which you attract the earth.

A man is standing on the surface of earth.$\\$ The force acting on the man = mg.....(I)$\\$ Assuming that, m = mass of the man = 50 Kg $\\$ and $\\$ g = accelaration due to gravity on the surface of the earth = 10 $\frac{m}{s^2}$ $\\$ W = mg = 50 x 10 = 500 N = force acting on the man $\\$ So, the man is also attracting the earth with a force of 500 N.

3.   3. At what distance should two charges, each equal to $1$ $C$, be placed so that the force between them equals your weight ?

The force of attraction between the two charges = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1q_2}{r^2}$ = 9 x ${10^9} $ $\frac{1}{r^2}$ $\\$ The force of attraction is equal to the weight. $\\$ $M_{g}$ = $\frac{9\times 10^9}{r^2}$ $\\$ $\Rightarrow$ ${r^2}$ = $\frac{9\times {10^9}}{m\times10}$ = $\frac{9\times10^8}{m}$ [taking g = 10 m/${s^2}$] $\\$ $\Rightarrow$ r = $\sqrt\frac{9\times{10^8}}{m}$ = $\frac{3\times{10^4}}{\sqrt m}$ mt $\\$ For example,Assuming m = 64 kg. $\\$ r = $\frac{3\times{10^4}}{\sqrt{64}} $= $\frac{3}{8}$ ${10^4}$ = 3750 m.

4.   4. Two spherical bodies, each of mass $50$ kg, are placed at a separation of $20$ cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.

mass = 50kg.$\\$ r = 20 cm = 0.2m$\\$ ${F_G}$ = G$\frac{m_1m_2}{r^2}$ =$\frac {6.67\times10^{-11}\times2500}{0.04}$ $\\$ Coulomb's force ${F_c}$ = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1q_2}{r^2}$ =$\frac {9\times 10^{9}\times q^{2}}{0.04}$ $\\$ Since, ${F_g}$ = ${F_c}$ = $\frac {6.7\times10^{-11}\times2500}{0.04}$ = $\frac {9\times 10^{9}\times q^{2}}{0.04}$ $\\$ $\Rightarrow$ $q^2$ = $\frac {6.7\times10^{-11}\times2500}{0.04}$ =$\frac{6.67\times10^{-9}}{9\times10^9}$ x 25 $\\$ = 18.07 x $10^{-18}$ $\\$ q = $\sqrt{18.07\times10^{-18}}$ = 4.3 x $10^{-9}c$

5.   5. A monkey is sitting on a tree limb. The limb exerts a normal force of $48 N$ and a frictional force of $20 N$. Find the magnitude of the total force exerted by the limb on the monkey.

The limb exerts a normal force $48 N$ and frictional force of $20 N$.Resultant magnitude of the force.$\\$ R = $\sqrt{(48)^2 + (20)^2}$ $\\$

= $\sqrt{2304 + 400}$ $\\$ = $\sqrt{2704}$ $\\$ = $52 N.$

6.   6. A body builder exerts a force of $150 N$ against a bullworker and compresses it by $20$ cm. Calculate the spring constant of the spring in the bullworker.

At h = $36000 km.g^{1}$ = $\frac{GM}{(36000+6400)^2}$

The body builder exerts a force = $150 N.$ $\\$ Compression x = $20 CM$ = $0.2 m$ $\\$ $\therefore$ Total force exerted by the man = f = kx. $\\$ $\Rightarrow$ Kx = 150 $\\$ $\Rightarrow$ K=$\frac{150}{0.2} $= $\frac{1500}{2}$ = 750$\frac{N}{m}$

7.   7. A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station ? (Radius of the earth is $6400$ km.)

Suppose the height is h.$\\$ At earth station F = $\frac{GMm}{R^2}$ $\\$ M = mass of earth.$\\$ m=mass of satelite.$\\$ R = radius of earth.$\\$ F = $\frac{GMm}{(R + h)^2}$ = $\frac{GMm}{2R^2}$ $\\$ $\Rightarrow 2R^2$ = $(R+h)^2$ $\Rightarrow R^2 - h^2 - 2Rh $= 0 $\\$ $\Rightarrow h^2 + 2Rh - R^2$ = 0 $\\$ H = $\frac{(-2R\pm\sqrt{4R^2 + 4R^2})}{2}$ = $\frac{-2R\pm2\sqrt{2R}}{2}$$\\$ =$ -R\pm \sqrt{2R}$ = $R(\sqrt2 - 1)$$\\$ = $6400\times(0.414)$$\\$ =2649.6 = 2650 km.

8.   8. Two charged particles placed at a separation of $20 cm$ exert $420 N$ of Coulomb force on each other. What will be the force if the separation is increased to $25 cm$ ?

Two charged particle placed at a sehortion 2m. exert a force of 20m.$\\$ F1 = 20 N.$\\$ F2 = ?$\\$ r1 = 20 cm. $\\$ r2 = 25 cm.$\\$ Since, F = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1q_2}{r^2}$, $\\$ F = $\frac{1}{r^2}$ $\\$ $\frac{f_1}{f_2}$ = $\frac{r^2_2}{r^2_1}$ $\Rightarrow F_2$ = $F_1$ x$\big(\frac{r_1}{r_2}\big)^2 $ = 20 x $\big(\frac{20}{25}\big)^2 $ = 20 x $\frac{16}{25}$ = $\frac{64}{5}$ = 12.8 N = 13 N. Two charged particle placed at a sehortion 2m. exert a force of 20m.$\\$ F1 = 20 N.$\\$ F2 = ?$\\$ r1 = 20 cm. $\\$ r2 = 25 cm.$\\$ Since, F = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1q_2}{r^2}$, $\\$ F = $\frac{1}{r^2}$ $\\$ $\frac{f_1}{f_2}$ = $\frac{r^2_2}{r^2_1}$ $\Rightarrow F_2$ = $F_1$ x$\big(\frac{r_1}{r_2}\big)^2 $ = 20 x $\big(\frac{20}{25}\big)^2 $ = 20 x $\frac{16}{25}$ = $\frac{64}{5}$ = 12.8 N = 13 N.

9.   9. The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data : The universal constant of gravitation $G = 6.67 $ x $10-11 N-m 2/kg 2$, mass of the $moon$ = $7.36$ x $1022 kg$, mass of the $earth = 6$ x $10 24 kg$ and the distance between the earth and the $moon = 3.8$ x $10 5 km.$

The force between the earth and the moon ,$\\$ F = G $\frac{m_m m_c}{r^2}$$\\$ F =$\frac{6.67\times10^{-11}\times7.36\times10^{22}\times6\times10^{24}}{(3.8\times10^8)^2}$$\\$ = $\frac{6.67\times7.36\times\times10^35}{(3.8)^2\times10^{16}}$$\\$ = $20.3\times10^{19}$ = $2.03\times10^{20} N$ = $2\times10^{20} N$.

10.   10. Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

Charge on portion = $1.6\times10^{-19}$ $\\$ $\therefore$ $ F_{electrical}$ = $\frac{1}{4\pi\varepsilon_0}$ x $\frac{q_1 q_2}{r^2}$ = $\frac{9\times10^9\times(1.6)^2\times10^{-38}}{r^2}$ $\\$ mass of proton = $1.732\times10^{-27}$ Kg$\\$ $F_{Gravity}$ = G $\frac{m_1m_2}{r^2}$ = $\frac{6.67\times10^{-11}\times(1.732)\times10^{-54}}{r^2}$ $\\$ $\frac{F_e}{F_g}$ = $\frac{9\times(1.6)^{2}\times10^{-29}}{6.67(1.732)^{2}10^{-65}}$ $\\$= $1.24\times10^{36}$

11.   11. The average separation between the proton and the electron in a hydrogen atom in ground state is $5.3$ x $10 - 11 $i n. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state ?

The average separation between proton and electron of hydrogen atom is = 5.3 x $10^{-11}m$.$\\$ a) Coulomb's force F = 9 x $10^9$ x $\frac{q_1q_2}{r^2}$ = $\frac{9\times10^{9} \times (1.0\times 10^{-19})^{2}} {(5.3 \times10^{-11})^2}$ = 8.2 x $10^{-8}N$.$\\$ b) When the average distance between proton and electron becomes 4times that of its ground state.$\\$ Coulomb's force F =$\frac{1}{4\pi\varepsilon_0}$ x $\frac{q_1q_2}{(4r)^2}$ = $\frac{9 \times10^{9} \times(1.6 \times10^{-19})^{2}} {16 \times(5.3)^{2}\times10^{-22}}$ $\\ $ = $\frac{9\times(1.6)^{2}}{16\times(5.3)^{2}}$ x $10^{-7}$ $\\$ =$ 0.0512\times10^{-7} $ = $ 5.1 \times10^{-9}N.$

12.   12. The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a $120-kg$ equipment placed in a geostationary satellite. The radius of the earth is $6400 km$.

The geostationary orbit of earth is at distance of about $36000 km$.$\\$ We know that, $g^{1}$ = $\frac{GM}{(R+h)^2}$$\\$ At h = $36000 km.g^{1}$ = $\frac{GM}{(36000+6400)^2}$ $\\$ $\therefore$ $\frac{g^{1}}{g}$ = $\frac{6400\times6400}{42400\times42400}$ = $\frac{256}{106\times106}$ = 0.0227$\\$ $\Rightarrow$$g^{1}$ = 0.0227 x 9.8 = 0.223$\\$ [ taking g = 9.8 $\frac{m}{s^{2}}$ at the surface of the earth]$\\$ A$120 kg$ equipment placed in a geostationary satellite will have weight $\\$ Mg = 0.233 x 120 = 26.79 = $27N$

13.   Thanks

Two charged particle placed at a sehortion 2m. exert a force of 20m.$\\$ F1 = 20 N.$\\$ F2 = ?$\\$ r1 = 20 cm. $\\$ r2 = 25 cm.$\\$ Since, F = $\frac{1}{4\pi\varepsilon_0}$ $\frac{q_1q_2}{r^2}$, $\\$ F = $\frac{1}{r^2}$ $\\$ $\frac{f_1}{f_2}$ = $\frac{r^2_2}{r^2_1}$ $\Rightarrow F_2$ = $F_1$ x$\big(\frac{r_1}{r_2}\big)^2 $ = 20 x $\big(\frac{20}{25}\big)^2 $ = 20 x $\frac{16}{25}$ = $\frac{64}{5}$ = 12.8 N = 13 N.

14.   Thanks

The force between the earth and the moon ,$\\$ F = G $\frac{m_m m_c}{r^2}$

F =$\frac{6.67\times10^{-11}\times7.36\times10^{22}\times6\times10^{24}}{(3.8\times10^8)^2}$

= $\frac{6.67\times7.36\times\times10^35}{(3.8)^2\times10^{16}}$

= $20.3\times10^{19}$ = $2.03\times10^{20} N$ = $2\times10^{20} N$.

15.   Thanks

Charge on portion = $1.6\times10^{-19}$ $\\$ $\therefore$ $ F_{electrical}$ = $\frac{1}{4\pi\varepsilon_0}$ x $\frac{q_1 q_2}{r^2}$ = $\frac{9\times10^9\times(1.6)^2\times10^{-38}}{r^2}$ $\\$ mass of proton = $1.732\times10^{-27}$ Kg$\\$ $F_{Gravity}$ = G $\frac{m_1m_2}{r^2}$ = $\frac{6.67\times10^{-11}\times(1.732)\times10^{-54}}{r^2}$ $\\$ $\frac{F_e}{F_g}$ = $\frac{9\times(1.6)^{2}\times10^{-29}}{6.67(1.732)^{2}10^{-65}}$ $\\$= $1.24\times10^{36}$

16.   Thanks

a) Coulomb's force = F =9 x $10^9$ x $\frac{q_1q_2}{r^2}$ $\\$

The average separation between proton and electron of hydrogen atom is = 5.3 x $10^{-11}m$.$\\$ a) Coulomb's force F = 9 x $10^9$ x $\frac{q_1q_2}{r^2}$ = $\frac{9\times10^{9} \times (1.0\times 10^{-19})^{2}} {(5.3 \times10^{-11})^2}$ = 8.2 x $10^{-8}N$.$\\$ b) When the average distance between proton and electron becomes 4times that of its ground state.$\\$ Coulomb's force F =$\frac{1}{4\pi\varepsilon_0}$ x $\frac{q_1q_2}{(4r)^2}$ = $\frac{9 \times10^{9} \times(1.6 \times10^{-19})^{2}} {16 \times(5.3)^{2}\times10^{-22}}$ $\\ $ = $\frac{9\times(1.6)^{2}}{16\times(5.3)^{2}}$ x $10^{-7}$ $\\$ =$ 0.0512\times10^{-7} $ = $ 5.1 \times10^{-9}N.$