JEE Main · PhysicsMedium

$A block of mass m is attached to a spring of spring constant k and is pulled from its equilibrium position by a distance x . If the block is released from rest, what is the work done by the spring force when the block moves from x = A to x = A /2 ?$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{3}{8} k A^{2}$

Solution

$The work done by the spring is W = - Δ U = - (\frac{1}{2} k x_{f}^{2} - \frac{1}{2} k x_{i}^{2}) .$
$Here x_{i} = A and x_{f} = A /2 .$
$W = - \frac{1}{2} k ((A /2)^{2} - A^{2}) .$
$W = - \frac{1}{2} k (A^{2} /4 - A^{2}) = - \frac{1}{2} k (- 3 A^{2} /4) .$
$Hence the answer is (A).$