The backside of a truck is open and a box of 40 kg is placed 5 m away from the rear end. The coefficient of friction of the box with the surface of the truck is 0.15. The truck starts from rest with $2\;{\rm{m}}/{{\rm{s}}^2}$ acceleration. Calculate the distance covered by the truck when the box falls off.

The various forces acting on the block are as shown As the truck moves in forward direction with acceleration $2\;{\rm{m}}/{{\rm{s}}^2}$, the box experiences a force F , in backward direction, $F = ma = 40 \times 2 = 80N$ , in backward direction. Its motion will be opposed by force of friction $f = {\rm{\mu }}\;N = {\rm{\mu }}\;mg = 0.15 \times 40 \times 10 = 60$ The acceleration of the box relative to the truck toward the rear end is $a = \frac{{F - f}}{m} = \frac{{80 - 60}}{{40}} = 0.5\;m/{s^2}$ If t be the time taken by the box to fall off the truck $s = ut + \frac{1}{2}a{t^2}\,,\,5 = 0 + \frac{1}{2} \times 0.5 \times {t^2}\,,\,t = \sqrt {20} \;s$ During this time distance covered by truck $x = 0 \times t + \frac{1}{2} \times 2 \times ({\sqrt {20)} ^2} = 20{\rm{\;m}}$

A block compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then the pressure in the compartment is

The pressure on the rear side would be more due to fictitious force (acting in the opposite direction of acceleration) on the rear face. Consequently the pressure in the front side would be lowered

The correct Option is "Lower in front side"

A boy of mass 0.25 kg is projected with muzzle velocity $100\;m{s^{ - 1}}$ from a tank of mass 100 kg. What is the recoil velocity of the tank

Using law of conservation of momentum, we get $100 \times v = 0.25 \times 100 \Rightarrow v = 0.25\;m/s$

The correct Option is "0.25 $m{s^{ - 1}}$"

In figure, the blocks A, B and C each of mass m have acceleration ${a_1},{a_2}\,and\,{a_3}$ respectively. ${F_1}\,and\,{F_2}$ are external forces of magnitude 2 mg and mg respectively.Then

$\begin{aligned} & \mathrm{a}_1=\frac{2 \mathrm{mg}-\mathrm{mg}}{\mathrm{m}}=\mathrm{g} \\ & \mathrm{a}_2=\frac{2 \mathrm{mg}-\mathrm{mg}}{3 \mathrm{~m}}=\mathrm{g} / 3 \\ & \mathrm{a}_3=2 \mathrm{mg}-\mathrm{mg} / 2 \mathrm{~m} =\mathrm{g} / 2\end{aligned}$

The correct Option is "${a_1} > {a_3} > {a_2}$"

An 80 kg person is parachuting and is experiencing a downward acceleration of $\;2.8{\rm{m}}{{\rm{s}}^{ - 2}}$. The mass of the parachute is 5 kg. The upward force on the open parachute is (Take $g = 9.8{\rm{m}}{{\rm{s}}^{ - 2}})$

Mass of the person M=80 kg Mass of the parachute m=5 kg $\therefore $ Total mass of the system =M+m=85kg Downward acceleration $a = 2.8\;{\rm{m}}{{\rm{s}}^{ - 2}}$ Let upward force =F Applying Newton,s II law of motion to this system $F = \left( {m + M} \right)g - \left( {m + M} \right)a\,{\rm{or\;\;}}F = \left( {m + M} \right)\left( {g - a} \right)$ , $F = 85\left( {9.8 - 2.8} \right){\rm{N}} = 85 \times 7N,\,F = 595\,N$

The time in which a force of 2 N produces a change in momentum of $0.4\;kg - m{s^{ - 1}}$ in the body is

$\vec F = \frac{{\Delta \vec p}}{{\Delta t}} \Rightarrow \Delta t = \frac{{\left| {\Delta \vec p} \right|}}{{\left| {\vec F} \right|}} = \frac{{0.4}}{2} = 0.2\;s$

A gardner waters the plants by a pipe of diameter 1mm. The water comes out at the rate or $10\;c{m^3}/sec$. The reactionary force exerted on the hand of the gardner is

Rate of flow of water $\frac{V}{t} = \frac{{10\;c{m^3}}}{{{\rm{sec}}}} = 10 \times {10^{ - 6}}\frac{{{m^3}}}{{{\rm{sec}}}}$ Density of water $\rho = \frac{{{{10}^3}kg}}{{{m^3}}}$ Cross-sectional area of pipe $A = \pi {\left( {0.5 \times {{10}^{ - 3}}} \right)^2}$ Force $ = m\frac{{dv}}{{dt}} = \frac{{mv}}{t} = \frac{{V\rho v}}{t} = \frac{{\rho V}}{t} \times \frac{V}{{At}} = {\left( {\frac{V}{t}} \right)^2}\frac{\rho }{A}\;\left( {v = \frac{V}{{At}}} \right)$ By substituting the value in the above formula we get F=0.127 N

The correct Option is "$0.127\;N$"

When forces${F_1},\;{F_2},\;{F_3}$ are acting on a particle of mass m such that ${F_2}\,and\,{F_3}$ are naturally perpendicular, then the particle remains stationary. If the force ${F_1}\,$ is now removed then the acceleration of the particle is

Force equilibrium of system, ${F_1} = \sqrt {F_2^2 + F_3^2} $ As $\theta = 90^\circ $] In the absence of force ${F_1}$, Acceleration $ = \frac{{{\rm{Net\;Force}}}}{{{\rm{Mass}}}}\,,\, = \frac{{\sqrt {F_2^2 + F_3^2} }}{m} = \frac{{{F_1}}}{m}$

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