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

$$2\,T\cos {60^ \circ } = mg\,\,{\text{or}}\,\,T = mg = 2 \times 10 = 20\,N.$$

$$T$$ = Tension caused in string by monkey
$$ = m\left( {g + a} \right)$$
$$\eqalign{ & = m\left( {g + a} \right) \cr & \therefore T \leqslant 25 \times 10 \Rightarrow 20\left( {10 + a} \right) \leqslant 250 \cr & {\text{or,}}\,10 + a12.5 \Rightarrow a \leqslant 2.5 \cr} $$

$$F = \frac{{m\left( {v - u} \right)}}{t} = \frac{{0.15\left( {0 - 20} \right)}}{{0.1}} = 30\,N$$

Mass of man $$M = 80\,kg$$
Acceleration of lift, $$a = 5\,m/{s^2}$$
When, lift is moving upwards, the reading of weighing scale will be equal to $$R.$$
The equation of motion gives $$R - Mg = Ma$$
$$\eqalign{ & {\text{or}}\,R = Mg + Ma = M\left( {g + a} \right) \cr & \therefore R = 80\left( {10 + 5} \right) \cr & = 80 \times 15 = 1200\,N \cr} $$

In the case of masses hanging from a pulley by a string, the tension in whole string is same, say equal to $$T.$$

As $${M_2} > {M_1},$$   so mass $${M_2}$$ moves down and mass $${M_1}$$ moves up with the same acceleration a (say). The arrangement of the motion is represented in the figure. According to free body diagram of mass $${M_2},$$ is
$${M_2}g - T = {M_2}a\,......\left( {\text{i}} \right)$$
According to free body diagram of mass $${M_1},$$ is
$$T - {M_1}g = {M_1}a\,......\left( {{\text{ii}}} \right)$$
Adding Eqs. (i) and (ii), we get
$$\eqalign{ & \left( {{M_2}g - T} \right) + \left( {T - {M_1}g} \right) = \left( {{M_1} + {M_2}} \right)a \cr & \left( {{M_2} - {M_1}} \right)g = \left( {{M_1} + {M_2}} \right)a \cr & \Rightarrow a = \left( {\frac{{{M_2} - {M_1}}}{{{M_1} + {M_2}}}} \right)g \cr} $$
Given, $${M_1} = 5\,kg,\,{M_2} = 10\,kg$$
Hence, $$a = \left( {\frac{{10 - 5}}{{5 + 10}}} \right)g$$
$$\eqalign{ & = \frac{5}{{15}}g \cr & = \frac{g}{3}m/{s^2} \cr} $$
Alternative
Alternative, $$a = \frac{{\left( {{F_{{\text{net}}}}} \right){\text{system}}}}{{{\text{Net}}\,{\text{mass}}}}$$
$$\eqalign{ & = \frac{{\left( {10 - 5} \right) \times g}}{{5 + 10}} \cr & = \frac{g}{3}m/{s^2} \cr} $$
NOTE
In a mass-pulley system, the tension in the string is always towards the pulley.

Since, all the blocks are moving with constant velocity and we know that, if velocity is constant, acceleration of the body becomes zero. Hence, the net force on all the blocks will be zero.

Acceleration of system $$a = \frac{{{F_{{\text{net}}}}}}{{{M_{{\text{total}}}}}} = \frac{{14}}{{4 + 2 + 1}} = \frac{{14}}{7} = 2\,m/{s^2}$$

The contact force between $$A$$ and $$B$$
$$ = \left( {{m_B} + {m_C}} \right) \times a = \left( {2 + 1} \right) \times 2 = 6N$$

$$a = b + c$$
Net acceleration of $$A$$
$$\eqalign{ & = \sqrt {{a^2} + {c^2} + 2ac\cos \left( {\pi - \theta } \right)} \cr & = \sqrt {{{\left( {b + c} \right)}^2} + {c^2} - 2\left( {b + c} \right).c.\cos \theta } \cr} $$

Breaking strength is tension in the branch of tree. Let $$T$$ be tension in the branch of tree when monkey is descending with acceleration $$a.$$ Applying Newton's 2^nd law of motion
$$mg - T = ma$$
and $$T = 75\% $$   of weight of monkey i.e. $$T = \left( {\frac{{75}}{{100}}} \right)mg$$
$$\eqalign{ & = \frac{3}{4}mg \cr & \therefore mg - \frac{3}{4}mg = ma \cr & {\text{or}}\,a = \frac{g}{4} \cr} $$

The acrobat has a force acting on his hand that we resolve into two perpendicular components: the vertical one is the reaction to the weight $$\left( {115 \times 9.8\,N = 1127\,N} \right)$$     and the horizontal one balances the $$130\,N$$  force from the wall. These two forces give a resultant force $$F$$ of
$$F = \sqrt {{{1127}^2} + {{130}^2}} = 1134\,N$$