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Standard XII

Physics

Velocity of COM

Two blocks of...

Question

Two blocks of mass m and 2m are connected as shown in figure. A force F is acting on the block of mass 4m. Pulley is massless and string is light and inextensible. All the surfaces are smooth.

Contact force between ground and the 4m mass under the previous problem condition is

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3mg

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5mg

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7mg

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Solution

The correct option is D
7mg

Considering vertical equilibrium,

N= N2 + 4mg + T

= 2mg + 4mg + mg

= 7mg

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Similar questions

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(P) Acceleration of mass $M_{1}$	(1) $\frac{F}{M_{2}}$
(Q) Acceleration of mass $M_{2}$	(2) $\frac{F}{M_{1} + M_{2}}$
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(S) Tension in the string	(4) $\frac{M_{2} F}{M_{1} + M_{2}}$

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In the arrangement shown in the figure, there is friction between the blocks of masses m and 2m which are in contact. The ground is smooth. The mass of the suspended block is m. The block of mass m which is kept on mass 2m is stationary with respect to a block of mass 2m. The force of friction between m and 2m is (pulleys and string are lig frictionless):

Q. Three blocks

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Two blocks of mass m and 2m are connected as shown in figure. A force F is acting on the block of mass 4m. Pulley is massless and string is light and inextensible. All the surfaces are smooth. Contact force between ground and the 4m mass under the previous problem condition is

The correct option is D 7mg Considering vertical equilibrium, N= N2 + 4mg + T = 2mg + 4mg + mg = 7mg

Two blocks of mass m and 2m are connected as shown in figure. A force F is acting on the block of mass 4m. Pulley is massless and string is light and inextensible. All the surfaces are smooth.

Contact force between ground and the 4m mass under the previous problem condition is

The correct option is D
7mg

Considering vertical equilibrium,

N= N2 + 4mg + T

= 2mg + 4mg + mg

= 7mg