Two blocks `A` and `B` of masses `1kg` and `2kg` have acceleration `(2hat(i))m//s^(2)` and `(-4hat(j))m//s^(2)`. Find
(a) Psseudo force on block `A` as applied with respect to the block `B`.
(a) Pseudo force on block `B` as applied with respect to the block `A`

Two blocks A and B of masses 1kg and 2kg have acceleration (2hat(i))m//s^(2) and (-4hat(j))m//s^(2) . Find (a) Pseudo force on block A as applied with respect to the block B . (b) Pseudo force on block B as applied with respect to the block A

A block of mass 2 kg is kept at rest on a big box moving with velocity 2hat(i) and having acceleration -3hat(i)+4hat(j)m//s^(2) . Find the value of ‘Pseudo force’ acting on block with respect to box

The metal blocks A and B of masses 2 kg and 1 kg , respectively are in contact on a frictionless table A constant horizontal force of 3 N is applied to the block A . Calculate the force of constant between the two blocks . If the same force is applied to the block B then what will be the force of contact ? Can you explain the difference ?

Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

A block of mass 5 kg is placed on bus moving with acceleration 2 m//s^(2) .pseudo force acting on block as seen by a man on ground is -

A block A is just placed inside a smooth box B as shown in figure. Now the box is given an acceleration a = (2 hat(i) - 2 hat(j)) m//s ^(-2) . Under the acceleration block A cannot remain in the position shown. Block will require ma force for moving with acceleration a .

Assertion: A block A is just placed inside a smooth box B as shown in figure. Now the box is given an acceleration a = (2 hat(i) - 2 hat(j)) m//s ^(-2) . Under the acceleration block A cannot remain in the position shown. Reason: Block will require ma force for moving with acceleration a .

A block of mass m kg is placed on a plank. The plank is moving with an acceleration 6 hat(i) + 6 hat(j) (in m//s^(2) ). The frictional force acting on the block is (mu = 0.5, g = 10 m//s^(2)) :

Three blocks A , B and C of masses 4kg , 2kg and 1kg respectively are in contact on a frictionless surface, as shown. If a force of 14N is applied on the 4kg block, then the contact force between A and B is.