Question

A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block when it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s

Answer 1

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

$mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s$ ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.