Question

Two identical blocks A and B are placed on two inclined planes as shown in Fig. 8.241. Neglect resistance and other friction. Read the following statements and choose options.
Statement I: The kinetic energy of A sliding to J will be greater than the kinetic energy of B on sliding to O.
Statement II: The acceleration of A will be greater than the acceleration of B when both are released on the inclined plane.
Statement III : The work done by the external agent to move the block slowly from position B to O is negative.

• A. Only statement I is true
• B. Only statement II is true
• C. Only I and III are true
• D. Only II and III are true

Answer 1

The correct option is D Only II and III are true

Since both are released from the same height $h$

$\Delta P.E_A=\Delta P.E_B=mgh$

This change is potential energy is equal to change in the kinetic energy of the blocks

$⟹\Delta K.E_A=\Delta K.E_B$

The acceleration of both blocks along the incline plane is :

$a_A=gsin{\theta }_A⟹a_A=g\times \frac{h}{\sqrt{h^2+l^2}}$

And $a_B=gsin{\theta }_B⟹a_B=g\times \frac{h}{\sqrt{h^2+(2l{)}^2}}$

So, $a_A>a_B$

When an external agent brings the block form B to O, it should apply a force pointing up the incline to balance $mg\sin {\theta }_B$ and the displacement in down the incline.

So $W_{ext}=F.s\cos {180}^0=-Fs...............$(here F and s are positive)

Hence the work by external agent is negative.

Therefore Option D is correct