1

Question

Two trolleys 1 and 2 are moving with accelerations a1 and a2, respectively, in the same direction. A block of mass m on trolley 1 is in equilibrium from the frame of observer stationary with respect to trolley 2. The magnitude of friction force on block due to trolley is (assume that no horizontal force other than friction force is acting on block).

Open in App

Solution

The correct option is **B** ma2

R.E.F image

∙ Let the acceleration of block of mass with respect to

the observer on trolley 2 be denoted as →abo

∙ It is given that the block of mass m is in equilibrium with respect to the observer on trolley 2.

⇒→abo=0

⇒→ab−→ao=0 (where →ab denotes the acceleration of block with respect to ground and →ao denotes the acceleration of observer

with respect to ground )

⇒→ab=→ab

as →ao=→a2⇒→ab=→a2

⇒ acceleration of block with respect to ground is

→a2

The friction force on block due to trolley is responsible

for the acceleration →a2

Free body diagram of block

∴f=ma2

∴ The magnitude of friction force on block due to

trolley is f=ma2

1

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program