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 block B is pushed momentarily along a horizontal surface with an initial velocity V. If is the coefficient of sliding friction between B and the surface, block B will come to rest after a time
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Friction is the retarding force for the block
F = ma = µR = µmg
Therefore, from the first equation of motion
v = u – at
0 = V – µg × t ⇒ V/μg = tCorrect Option: D
Friction is the retarding force for the block
F = ma = µR = µmg
Therefore, from the first equation of motion
v = u – at
0 = V – µg × t ⇒ V/μg = t
- A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction µ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is
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a = F - μR = 100 - 0.5 × (10 × 10) = 5 ms-2 m 10
Correct Option: C
a = F - μR = 100 - 0.5 × (10 × 10) = 5 ms-2 m 10
- A block of mass 1 kg is placed on a truck which accelerates with acceleration 5m/s². The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is
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Maximum friction force = µmg
= .6 × 1 × 9.8 = 5.88 N
But here required friction force
= ma = 1 × 5 = 5 NCorrect Option: A
Maximum friction force = µmg
= .6 × 1 × 9.8 = 5.88 N
But here required friction force
= ma = 1 × 5 = 5 N
- A person slides freely down a frictionless inclined plane while his bag falls down vertically from the same height. The final speeds of the man (VM) and the bag (VB) should be such that
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As there is only gravitational field which works.
We know it is conservative field and depends only on the end points. So, VM = VBCorrect Option: B
As there is only gravitational field which works.
We know it is conservative field and depends only on the end points. So, VM = VB
