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Laws of Motion MCQs for NEET — Physics Questions with Answers

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When two surfaces are coated with a lubricant, then they

Stick to each other

Slide upon each other

Roll upon each other

None of these

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Explanation

Surfaces always slide over each other.

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A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is

0.38

0.44

0.52

0.60

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Explanation

Coefficient of friction $μ_{s} = \frac{F_{l}}{R} = \frac{75}{m g} = \frac{75}{20 \times 9 .8} = 0 .38$

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The maximum speed that can be achieved without skidding by a car on a circular unbanked road of radius R and coefficient of static friction μ, is

$μ R g$

$R g \sqrt{μ}$

$μ \sqrt{R g}$

$\sqrt{μ R g}$

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Explanation

In the given condition the required centripetal force is provided by frictional force between the road and tyre.

$\frac{m v^{2}}{R} = μ m g$

∴ $v = \sqrt{μ R g}$

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A car is moving along a straight horizontal road with a speed v₀. If the coefficient of friction between the tyres and the road is μ, the shortest distance in which the car can be stopped is

$\frac{v_{0}^{2}}{2 μ g}$

$\frac{v_{0}}{μ g}$

${(\frac{v_{0}}{μ g})}^{2}$

$\frac{v_{0}}{μ}$

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Explanation

Retarding force $F = m a = μ R = μ m g$

∴ $a = μ g$

Now from equation of motion $v^{2} = u^{2} - 2 a s$

$\Rightarrow 0 = u^{2} - 2 a s$

⇒ $s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μ g}$

∴ $= \frac{v_{0}^{2}}{2 μ g}$

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A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull acting at an angle of 30° to the upward drawn vertical which causes the block to just slide is

29.43 N

219.6 N

21.96 N

294.3 N

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Explanation

For a block of mass 50 kg on a rough horizontal surface with coefficient of friction Î¼ = 0.6, the least force F required to cause sliding at an angle of 30Â° is given by F = Î¼N/sin(30Â°), where N is the normal force (equal to mg). Substituting the values, we get F = 0.6 Ã— 50 Ã— 9.8/0.5 = 294.3 N. This force must overcome the maximum static friction to initiate sliding.

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Assuming the coefficient of friction between the road and tyres of a car to be 0.5, the maximum speed with which the car can move round a curve of 40.0 m radius without slipping, if the road is unbanked, should be

25 m/s

19 m/s

14 m/s

11 m/s

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Explanation

$v = \sqrt{μ g r} = \sqrt{0 .5 \times 9 .8 \times 40} = \sqrt{196} = 14 m / s$

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Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g = 10 m/s²)

30 m

40 m

72 m

20 m

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Explanation

$s = \frac{u^{2}}{2 μ g} = \frac{{(20)}^{2}}{2 \times 0 .5 \times 10} = 40 m$

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On the horizontal surface of a truck (μ = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at the rate of 5m/sec² then frictional force on the block will be

5 N

6 N

5.88 N

8 N

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Explanation

$F_{l} = μ m g = 0 .6 \times 1 \times 9 .8 = 5 .88 N$

Pseudo force on the block = $m a = 1 \times 5 = 5 N$

Pseudo is less then limiting friction hence static force of friction = 5 N.

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A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be μ, then the stopping distance is

$\frac{P}{2 μ m g}$

$\frac{P^{2}}{2 μ m g}$

$\frac{P}{2 μ m^{2} g}$

$\frac{P^{2}}{2 μ m^{2} g}$

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Explanation

$S = \frac{u^{2}}{2 μ g} = \frac{m^{2} u^{2}}{2 μ g m^{2}} = \frac{P^{2}}{2 μ m^{2} g}$

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Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is $[μ_{k} = 0 .5]$

100 m

400 m

800 m

1000 m

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Explanation

$s = \frac{u^{2}}{2 μ g} = \frac{{(100)}^{2}}{2 \times 0 .5 \times 10} = 1000 m$

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