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Laws of Motion Part 1

Section: Physics

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Question : 38 of 100

Marks: +1, -0

The boxes of masses

2 \text{kg}

8 \text{kg}

are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass

8 \text{kg}

to strike the ground starting from rest.

g=10 \text{m}/\text{s}^{2}

[27 Aug 2021 Shift 2]

0.34 s
0.2 s
0.25 s
0.4 s

Solution:

According to the given figure, and free body diagram

masses of two bodies

m_{1}=2 \text{kg} m_{2}=8 \text{kg}

,

Acceleration of mass

m_{1}=a

Acceleration of mass

m_{2}=\;\frac{a}{2}

Tension

=T

Distance between

m_{2}

and ground

=20 \text{cm}

=0.2 \text{m}

Initial velocity

u=0

Equation of motion of

2 \text{kg}

block,

\therefore \;\; T-2g=2a \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

Equation of motion of

8 \text{kg}

block, and

8g-2T=8 \;\frac{a}{2} \Rightarrow 4g-T=2a \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

From Eqs. (i) and (ii),

\Rightarrow \;\; T-2g=4g-T

\; \text{Substituting the value in Eq. (i) we get} \;

\;\; 3g-2g=2a \Rightarrow g=2a

\Rightarrow \;\; a=\;\frac{g}{2}=5 \text{ms}^{-2}

\therefore \;\; a_{1}=5 \text{ms}^{-2} \; \text{and} \; a_{2}=\;\frac{5}{2} \text{ms}^{-2}

\; \text{Since,} \; s=ut+\;\frac{1}{2} a t^{2}

\therefore \;\; \frac{20}{100}=0+\;\frac{1}{2} \;\frac{5}{2} \times t^{2} \Rightarrow t^{2}=\;\frac{20 \times 4}{5 \times 100}

\Rightarrow \;\; t=\;\frac{2}{5}=0.4 \text{s}

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