JEE Main 2022 · 25 Jun · Shift 1 · Physics Q52

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PhysicsInteger+4 / -1Question 52 of 90

A 0.5 kg block moving at a speed of $12\ \mathrm{ms^{-1}}$ compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be ______ $\mathrm{Nm^{-1}}$ .

Your answer

Answer and worked solution

Official answer: 600

Energy. $\frac12m\left(v^2-\frac{v^2}4\right)=\frac12kx^2$ : $\frac12(0.5)(144-36)=27$ J $=\frac12k(0.3)^2$ .

So $k=600$ N m $^{-1}$ .

Answer: 600

Topics:Spring potential energy Work-energy theorem

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JEE Main 2022 · 25 Jun · Shift 1, Physics question 52