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Laws of Motion MCQs for NEET — Physics Questions with Answers

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On the horizontal surface of a truck ( = 0.6 ) a block of mass 1 kg is placed. If the truck is accelerating at the rale of $ 5 m / s^ 2 $ then frictional force on the block will be_________ N

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Explanation

The frictional force can be found using the formula: \( f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, the normal force \( N \) is equal to the weight of the block, \( mg \). Here, \( \mu = 0.6 \), \( m = 1 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \). Thus, \( N = 1 \ \text{kg} \times 9.8 \ \text{m/s}^2 = 9.8 \ \text{N} \). The frictional force is then \( f = 0.6 \times 9.8 \ \text{N} = 5.88 \ \text{N} \). However, the maximum frictional force is \( \mu mg \), and if the truck is accelerating at \( 5 \ \text{m/s}^2 \), the frictional force required to prevent slipping is \( f = ma = 1 \ \text{kg} \times 5 \ \text{m/s}^2 = 5 \ \text{N} \). Therefore, the correct answer is 5 N.

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A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, the minimum radius of the arc at which the car turns is ________ meter.

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Explanation

$ v = \sqrt { rg } $ $ \therefore r = { v^2 / g } $

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A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time to if the lift is stationary and the time $t_2$ if it is accelerated in upward direction. Than

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Explanation

$ d = { 1 \over 2 } gt_1^2 and \, d = {1 \over 2 } (g + a ) t_2^2 $ By comparing both eqn $ {1 \over 2 } gt_1^2 = { 1 \over 2 } (g+ a)t_2^2 $ $ \therefore t_1 \gt t_2 $

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A lift of mass 1000 kg is moving with an acceleration of $1 ms^{–2}$ in upward direction Tension developed in the rope of lift is__________ N $ (g = 9.8 ms^{-2} )$

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Explanation

Tensile force T = m(g+a )

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A rope which can withstand a maximum tension of 400 N hangs from a tree. If a monkey of mass 30 kg climbs on the rope in which of the following cases-will the rope break? (take $g =10 ms^{-2} $ and neglect the mass of rope)

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Explanation

Calculate F - m(g+a)

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Same forces act on two bodies of different mass 2 kg and 5 kg initialy at rest. The ratio of times required to acquire same final velocity is _

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Explanation

$ Ft_1 = m_1 ( \nu -\nu_0) = m_1 \nu_1 $ $ Ft_2 = m_2 ( \nu -\nu_0) = m_2 \nu$ $ \therefore { t_1 \over t_2 } = { m_1 \over m_2 } $

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A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of $ 4 ^\circ $ with the vertical, The acceleration of the train is ___ $ms^{-2} (g = 10 ms^{-2} ) $

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Explanation

$ ma = mg sin \theta $ $ \therefore a = 10 sin 4 ^\circ $ $ \therefore a = 0.7 ms^{-2} $

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A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium The coefficient of friction between the chain and table is

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Explanation

= M/ L Weight of chain of lenght l = lg Weight of a chain of lenght ( L -l ) = (L -l ) g $ f_S = s_ R = s (L- l) g and f_S = l g $ $ \therefore l g = s. ( L - l ) g $ $ \therefore s = { l \over L - l } $

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A car of mass 1000 kg travelling at 32 m/s clashes into a rear of a truck of mass 8000 kg moving in the same direction with a velocity of 4 m/s. After the collision the car bounces with a velocity of $8 ms^{–1} $. The velocity of truck after the impact is m/s

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Explanation

According to eqn $ m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec v_1 + m_2 \vec v_2 $ $ \therefore v_2^1 = 9 m/s $

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The upper half of an inclined plane of inclination $ \theta $ is perfectly smooth while the lower half is rough A body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by_ ____

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Explanation

For upper half $ v_1^2 - v_o^2 = 2 ad $ $ \nu^2 - 0 = 2 ( g sin \theta )_2 $ $ \therefore v^2 = gl sin \theta $ For lower half $ v_1^2 - v_0^2 = 2 a' d $ $ 0 - gl sin \theta = 2g [ sin \theta - \mu_s cos \theta ] . { l \over 2 } $ $ \therefore 2 = 2 tan \theta $

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