Laws of Motion MCQ Questions & Answers in Basic Physics | Physics

According to second law of motion, magnitude of force can be calculated by multiplying mass of the body and the acceleration produced in it.
$$\eqalign{ & {\text{or force }}F = ma \cr & {\text{Here, }}F = 10{\text{ }}N \cr & a = 1\,m/{s^2} \cr & \therefore m = \frac{F}{a} = \frac{{10}}{1} = 10\,kg \cr} $$

$$\eqalign{ & \frac{m}{2}g - T = \frac{m}{2}a\,......\left( {\text{i}} \right) \cr & T\cos {60^ \circ } = \frac{{ma}}{{\cos {{60}^ \circ }}}\,......\left( {{\text{ii}}} \right) \cr} $$
Solving (i) and (ii)
acceleration of ring $$ = \frac{{2g}}{9}$$

$$\eqalign{ & F = \left( {m + m + m} \right) \times a \cr & \therefore a = \frac{{10.2}}{6}\,m/{s^2} \cr & \therefore {T_2} = ma = 2 \times \frac{{10.2}}{6} = 3.4\,N \cr} $$

$$\eqalign{ & mg - T = ma \cr & \therefore a = g - \frac{T}{m} = 10 - \frac{{360}}{{60}} = 4m/{s^2} \cr} $$

(Force diagram in the frame of the car) Applying Newton’s law perpendicular to string
$$mg\sin \theta = ma\cos \theta \Rightarrow \tan \theta = \frac{a}{g}$$
Applying Newton’s law along string
$$\eqalign{ & \Rightarrow T - m\sqrt {{g^2} + {a^2}} = ma \cr & {\text{or}}\,\,T = m\sqrt {{g^2} + {a^2}} + ma \cr} $$

$$ \bullet $$ For the man standing in the left, the acceleration of the ball
$${{\vec a}_{bm}} = {{\vec a}_b} - {{\vec a}_m} \Rightarrow {a_{bm}} = g - a$$
Where $$'a'$$ is the acceleration of the mass (because the acceleration of the lift is $$'a'$$ )
$$ \bullet $$ For the man standing on the ground, the acceleration of the ball
$${{\vec a}_{bm}} = {{\vec a}_b} - {{\vec a}_m} \Rightarrow {a_{bm}} = g - 0 = g$$

The velocity of parachutist when parachute opens is
$$u = \sqrt {2gh} = \sqrt {2'\,9.8'\,50} = \sqrt {980} $$
The velocity at ground, $$v = 3m/s$$
$$\therefore S = \frac{{{v^2} - {u^2}}}{{2 \times \left( { - 2} \right)}} = \frac{{{3^2} - 980}}{{ - 4}} \approx 243\,m$$
Initially he has fallen $$50m.$$
∴ Total height from where he bailed out $$ = 243 + 50 = 293m$$

Total mass $$\left( m \right) =$$  Mass of lift + Mass of person
$$ = 940 + 60 = 1000{\text{ }}kg$$
So, from the free body diagram
$$T - mg = ma$$

$$\eqalign{ & {\text{Hence,}}\,T - 1000 \times 10 = 1000 \times 1 \cr & T = 11000\,N \cr} $$

Let an acceleration to the wedge be given towards left, then the block (being in non inertial frame) has a pseudo acceleration to the right because of which the block is not slipping
$$\eqalign{ & \therefore mg\sin \theta = {a_{{\text{pseudo}}}}\cos \theta \cr & \Rightarrow {a_{{\text{pseudo}}}} = \frac{{mg\sin \theta }}{{\cos \theta }} \cr} $$

$$\eqalign{ & 2N\cos {30^ \circ } = W, \cr & N = \frac{W}{{\sqrt 3 }} \cr & {\text{and}}\,N\sin {30^ \circ } = F \cr & {\text{or}}\,\,\frac{W}{{\sqrt 3 }} \times \frac{1}{2} = F,\,{\text{or}}\,F = \frac{W}{{2\sqrt 3 }}. \cr} $$