The figure shows a mass m on a frictionless s | vedclass

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Question

(A) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{total}} = \Delta K$$W

Vedclass · Accepted Answer

$\sqrt{\frac{2 Fx - k x^2}{m}}$

Vedclass · Answer

(A) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.$W_{	ext{total}} = \Delta K$$W_F + W_{	ext{spring}} = K_f - K_i$Here,the work done by the constant force $F$ is $W_F = Fx$.The work done by the spring force is $W_{	ext{spring}} = -\frac{1}{2} k x^2$.The initial kinetic energy $K_i = 0$ (as it starts from rest) and the final kinetic energy $K_f = \frac{1}{2} m v^2$.Substituting these values into the equation:$Fx - \frac{1}{2} k x^2 = \frac{1}{2} m v^2 - 0$$Fx - \frac{1}{2} k x^2 = \frac{1}{2} m v^2$Multiplying both sides by $2$:$2Fx - k x^2 = m v^2$$v^2 = \frac{2Fx - k x^2}{m}$$v = \sqrt{\frac{2Fx - k x^2}{m}}$

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