Laws of Motion 2 Question 1
1. A block of mass $5 kg$ is (i) pushed in case (A) and (ii) pulled in case (B), by a force $F=20 N$, making an angle of $30^{\circ}$ with the horizontal, as shown in the figures. The coefficient of friction between the block, the floor is $\mu=0.2$. The difference between the accelerations of the block, in case (B) and case (A) will be (Take, $g=10 ms^{-2}$ )
(A)
(B) lifted up by attaching a mass $2 m$ to the other end of the rope. In Fig. (b), $m$ is lifted up by pulling the other end of the rope with a constant downward force $F=2 m g$. The acceleration of $m$ is the same in both cases.
$(1984,2 M)$
(a)
(b)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 1. (c)
Solution:
- Case I Block is pushed over surface
Free body diagram of block is
In this case, normal reaction,
$N=m g+F \sin 30^{\circ}=5 \times 10+20 \times \frac{1}{2}=60 N$
[Given, $m=5 kg, F=20 N$ ]
Force of function, $f=\mu N$
$$ \begin{aligned} & =0.2 \times 60 \quad[\because \mu=0.2] \\ & =12 N \end{aligned} $$
So, net force causing acceleration $\left(a _1\right)$ is
$$ F _{\text {net }}=m a _1=F \cos 30^{\circ}-f $$
$\Rightarrow \quad m a _1=20 \times \frac{\sqrt{3}}{2}-12$
$\therefore \quad a _1=\frac{10 \sqrt{3}-12}{5} \approx 1 ms^{-2}$
Case II Block is pulled over the surface
Free body diagram of block is,
Net force causing acceleration is
$$ \begin{aligned} & F _{\text {net }}=F \cos 30^{\circ}-f=F \cos 30^{\circ}-\mu N \\ \Rightarrow \quad & F _{\text {net }}=F \cos 30^{\circ}-\mu\left(m g-F \sin 30^{\circ}\right) \end{aligned} $$
If acceleration is now $a _2$, then
$$ \begin{aligned} & a _2=\frac{F _{\text {net }}}{m}=\frac{F \cos 30^{\circ}-\mu\left(m g-F \sin 30^{\circ}\right)}{m} \\ & =\frac{20 \times \frac{\sqrt{3}}{2}-0.25 \times 10-20 \times \frac{1}{2}}{5}=\frac{10 \sqrt{3}-8}{5} \\ \Rightarrow & a _2 \approx 1.8 ms^{-2} \end{aligned} $$
So, difference $=a _2-a _1=1.8-1=0.8 ms^{-2}$
