Laws of Motion

Given u = V, final velocity = 0.
Using v = u + at

$ \therefore $ $0 = V - at\,$

$ \Rightarrow $ $ - a = {{0 - V} \over t} = - {V \over t}$

f = $\mu $R = $\mu $mg (f is the force of friction)

$ \therefore $ Retardation, a = $\mu g$

So, $ t = {V \over a} = {V \over {\mu g}}$

The wedge is given an acceleration to the left.

$ \therefore $ The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.

$ \therefore $ mgsin$\theta $ = macos$\theta $

or $a = {{g\sin \theta } \over {\cos \theta }}$

Total reaction of the wedge on the block is N = mgcos$\theta $ + masin$\theta $.

$ \Rightarrow N = mg\cos \theta + {{mg\sin \theta \sin \theta } \over {\cos \theta }}$

$ \Rightarrow N = {{mg\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)} \over {\cos \theta }} = {{mg} \over {\cos \theta }}$

Free body diagram of two masses is



We get equations
$T + ma = {f_\mu }$ or $T = \mu {N_A}(for\,a = 0)$

and T = ma + mg or T = m_Bg (for a = 0)

$ \therefore $ $\mu {N_A} = {m_B}g$

$ \Rightarrow {m_B} = \mu {m_A} = 0.2 \times 2 = 0.4\,kg$

T = Tension caused in string by monkey

= m (g + a)

$T \le 25 \times 10 \Rightarrow 20\left( {10 + a} \right) \le 250$

$ \Rightarrow 10 + a \le 12.5$

$ \Rightarrow a \le 2.5$

Reading of the scale = Apparent wt. of the man = m(g + a)

= 80 (10 + 5) = 1200 N

m = 10 kg, R = mg

$ \therefore $ Frictional force = f_k

= ${\mu _k}R = {\mu _k}mg$

= 0.5 × 10 × 10

= 50 N [g = 10 m/sec²]

$ \therefore $ Net force acting on the body = F = P – f_k
= 100 – 50 = 50 N.

$ \therefore $ Acceleration of the block = a = F/m
= 50/10 = 5 m/sec² .

Mass (m) = 3 kg, force (F) = $6{t^2}\widehat i + 4t\widehat j$

$ \therefore $ acceleration
$a = F/m = {{6{t^2}\widehat i + 4t\widehat j} \over 3} = 2{t^2}\widehat i + {4 \over 3}t\widehat j$

Now, $a = {{dv} \over {dt}} = 2{t^2}\widehat i + {4 \over 3}t\widehat j$

$dv = \left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)dt$

$ \therefore v = \int\limits_0^3 {\left( {2{t^2}\widehat i + {4 \over 3}t\widehat j} \right)} dt$

= $\left. {{2 \over 3}{t^3}\widehat i + {4 \over 6}{t^2}\widehat j} \right|_0^3 = 18\widehat i + 6\widehat j$

For a lift which is moving in upward direction with an acceleration a, the tension T developed in the string connected to the lift is given by

T = m (g + a).

Here m = 1000 kg, a = 1 m/s², g = 9.8 m/s²

$ \therefore $ T = 1000(9.8 + 1) = 10,800 N.

Maximum friction force = $\mu $mg

= .6 × 1 × 9.8 = 5.88 N

But here required friction force

= ma = 1 × 5 = 5 N

Net force experienced = ${{Total\,{\mathop{\rm Impulse}\nolimits} } \over {Time\,taken}}$

$ = {{m\Delta v} \over t} = 0.15 \times {{20} \over {0.1}} = 30N$

Apply conservation of linear momentum. Total momentum before explosion = total momentum after explosion

$0 = {m \over 5}{v_1}\widehat i + {m \over 5}{v_2}\widehat j + {{3m} \over 5}{\overrightarrow v _3}$

${{3m} \over 5}{\overrightarrow v _3} = - {m \over 5}\left[ {{v_1}\widehat i + {v_2}\widehat j} \right]$

${\overrightarrow v _3} = {{ - {v_1}} \over 3}\widehat i - {{{v_2}} \over 3}\widehat j$
     $ \therefore $ ${v_1} = {v_2} = 30$ m/sec

${\overrightarrow v _3} = - 10\widehat i - 10\widehat j;{v_3} = 10\sqrt 2 $ m/sec

Load W = Mg = 75 × 10 = 750 N

Effort (P) = 250 N

$ \therefore $ Mechanical advantage
$ = {{load} \over {effort}} = {W \over P} = {{750} \over {250}} = 3$

Velocity ratio
$ = {{distance\,travelled\,by\,effort} \over {distance \,travelled\,by\,load}} = 4$

Efficiency $(\eta) = {{Mechanical\,advantage} \over {Velocity\,ratio}}$

$ = \left( {3/4} \right) \times 100 = 75\% $

Let T be the tension in the string.

$ \therefore $ 10g – T = 10a     ....(i)

T – 5g = 5a             ....(ii)

Adding (i) and (ii),

$5g = 15a \Rightarrow a = {g \over 3}m/{s^2}$

Change in momentum = 2 × 3 × 10 × sin60°

= $60 \times {{\sqrt 3 } \over 2}$

Force = Change in momentum/Impact time

$ = {{30\sqrt 3 } \over {0.2}} = 150\sqrt 3 N$