Laws of Motion
- A block of mass m is in contact with the cart C as shown in the Figure.
The coefficient of static friction between the block and the cart is . The acceleration of the cart that will prevent the block from falling satisfies:
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Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force mα due to acceleration
∴ N = mα
For the block not to fall, frictional force, Ff ≥ mg
⇒ μN ≥ mg
⇒ μmα ≥ mg
⇒ α ≥ g/μCorrect Option: C
Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force mα due to acceleration
∴ N = mα
For the block not to fall, frictional force, Ff ≥ mg
⇒ μN ≥ mg
⇒ μmα ≥ mg
⇒ α ≥ g/μ
- A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is µ = 0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 ms–2, is
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Frictional force on the box f = µmg
∴ Acceleration in the box
a = µg = 5 ms–2
v² = u² + 2as
⇒ 0 = 2² + 2 × (5) s
⇒ s = – 2/5 w.r.t. belt
⇒ distance = 0.4 mCorrect Option: D
Frictional force on the box f = µmg
∴ Acceleration in the box
a = µg = 5 ms–2
v² = u² + 2as
⇒ 0 = 2² + 2 × (5) s
⇒ s = – 2/5 w.r.t. belt
⇒ distance = 0.4 m
- The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
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For upper half of inclined plane
v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ
For lower half of inclined plane
0 = u² + 2 g (sin θ – µ cos θ) S/2
⇒ – gS sin θ = gS ( sinθ – µ cos θ)
⇒ 2 sin θ = µ cos θ
⇒ µ= 2sinθ/cosθ = 2 tan θCorrect Option: B
For upper half of inclined plane
v² = u² + 2a S/2 = 2 (g sin θ) S/2 = gS sin θ
For lower half of inclined plane
0 = u² + 2 g (sin θ – µ cos θ) S/2
⇒ – gS sin θ = gS ( sinθ – µ cos θ)
⇒ 2 sin θ = µ cos θ
⇒ µ= 2sinθ/cosθ = 2 tan θ
- A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope ? (g = 10 m/s²)
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T = Tension caused in string by monkey
= m (g + a)
∴ T ≤ 25 × 10 ⇒ 20(10 + a) ≤ 250
or, 10 + a ≤ 12.5 ⇒ a ≤ 2.5Correct Option: A
T = Tension caused in string by monkey
= m (g + a)
∴ T ≤ 25 × 10 ⇒ 20(10 + a) ≤ 250
or, 10 + a ≤ 12.5 ⇒ a ≤ 2.5
- A lift weighing 1000 kg is moving upwards with an accelertion of 1 m/s². The tension in the supporting cable is
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T – (1000 × 9.8 )= 1000 × 1
⇒ T = 10800 NCorrect Option: B
T – (1000 × 9.8 )= 1000 × 1
⇒ T = 10800 N
