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Byju's Answer

Standard XII

Physics

Newton's Second Law

In the shown ...

Question

In the shown figure $m_{A}, m_{B} a n d m_{C}$ are the masses of the three blocks, the incline is frictionless $m_{A} = 5 k g, m_{B} = 10 k g, m_{C} = 2 k g$ . coefficient of friction between A and B is 0.5 and between B and C is 0.1 (Take g = 10 $m / s^{2}$ ) On the basis of above data answer the following questions.

What is the frictional force on block A

30 N

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48 N

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18 N

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78 N

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Solution

The correct option is A
30 N

$m_{A g} s i n θ = 5 \times 10 \times s i n 37^{\circ} = 5 \times 10 \times \frac{3}{5} = 30 N f_{m a x} = μ (m_{C} + m_{B}) g c o s θ = 48 N Hence f = 30 N$

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Similar questions

In the shown figure $m_{A}, m_{B} a n d m_{C}$ are the masses of the three blocks, the incline is frictionless $m_{A} = 5 k g, m_{B} = 10 k g, m_{C} = 2 k g$ . coefficient of friction between A and B is 0.5 and between B and C is 0.1 (Take g = 10 $m / s^{2}$ ) On the basis of above data answer the following questions.

What is the tension in the rod ?

Q. Given

m_{A} = 30 k g, m_{B} = 10 k g, m_{C} = 20 k g

. Between

A

and

B

μ_{1} = 0.3

, between

B

and

C

μ_{2} = 0.2

and between

C

and ground

μ_{3} = 0.1

. The least horizontal force

F

to start motion of any part of the system of three blocks resting upon one another as shown in figure is

(g = 10 m / s^{2})

In the figure shown co-efficient of friction between the block B and the block C is 0.4. there is no friction between the block C and the surface on which it is placed. The system of blocks is released from rest in the situation shown in the diagram. Find the distance moved by the block C when block A descends through a distance 2 m. Given masses of the blocks are $m_{A}$ = 3 kg, $m_{B}$ = 5 kg and $m_{C}$ = 10 kg.

Q. Given the set up shown in the figure. Blocks

A, B

and

C

have masses

m_{A} = M

m_{B} = m_{C} = m

. The strings are assumed massless and unstretchable and the pulley is frictionless. There is no friction between block

B

and the support table, but there is friction between blocks

B

and

C

donated by a given coefficient

μ

. Find the accelaration of block

A

In the arrangement shown in ﬁgure $m_{A} = m_{B} = 2 k g$ . String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefﬁcient between blocks A and B is $μ = 0.5$ . What maximum horizontal force F can be applied so that block A does not slip over the block B? $(g = 10 m / s^{2})$

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The correct option is A 30 N mAgsin θ=5×10×sin 37∘=5×10×35=30Nfmax=μ(mC+mB)g cos θ=48NHence f = 30 N

The correct option is A
30 N

$m_{A g} s i n θ = 5 \times 10 \times s i n 37^{\circ} = 5 \times 10 \times \frac{3}{5} = 30 N f_{m a x} = μ (m_{C} + m_{B}) g c o s θ = 48 N Hence f = 30 N$