SATHEE: Laws of Motion 2 Question 1

Laws of Motion 2 Question 1

1. A block of mass $5 k g$ 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 $μ = 0.2$ . The difference between the accelerations of the block, in case (B) and case (A) will be (Take, $g = 10 m s^{- 2}$ )

(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, $m = 5 k g, F = 20 N$ ]

Force of function, $f = μ N$

$\begin{aligned} = 0.2 \times 60 [∵ μ = 0.2] \\ = 12 N \end{aligned}$

So, net force causing acceleration $(a_{1})$ is

$F_{net} = m a_{1} = F \cos 30^{\circ} - f$

$\Rightarrow m a_{1} = 20 \times \frac{\sqrt{3}}{2} - 12$

$∴ a_{1} = \frac{10 \sqrt{3} - 12}{5} \approx 1 m s^{- 2}$

Case II Block is pulled over the surface

Free body diagram of block is,

Net force causing acceleration is

$\begin{aligned} F_{net} = F \cos 30^{\circ} - f = F \cos 30^{\circ} - μ N \\ \Rightarrow & F_{net} = F \cos 30^{\circ} - μ (m g - F \sin 30^{\circ}) \end{aligned}$

If acceleration is now $a_{2}$ , then

$\begin{aligned} a_{2} = \frac{F_{net}}{m} = \frac{F \cos 30^{\circ} - μ (m g - F \sin 30^{\circ})}{m} \\ = \frac{20 \times \frac{\sqrt{3}}{2} - 0.25 \times 10 - 20 \times \frac{1}{2}}{5} = \frac{10 \sqrt{3} - 8}{5} \\ \Rightarrow & a_{2} \approx 1.8 m s^{- 2} \end{aligned}$