Question
A block of mass $2 \mathrm{~kg}$ is released from rest on a smooth inclined plane of inclination $30^{\circ}$ and connected by a massless spring of force constant $1000 \mathrm{~N} / \mathrm{m}$ as shown in Fig. $4.25$. Initially, the spring is in its natural length. An external variable force also acts on the block down the inclined plane. Block comes to rest for a momentafter travelling a distance of $20 \mathrm{~cm}$ along the inclined plane. From initial to this moment, Column-I gives work done by various forces and Column-II gives their values.(A) Work done by gravity(1) Zero(B) Work done by spring(2) $18 \mathrm{~J}$(C) Work done by external force(3) $-20 \mathrm{~J}$(D) Work done by normal force(4) $2 \mathrm{~J}$(5) $20 \mathrm{~J}$
Step 1
Here, m = 2 kg, g = 10 m/s$^2$, s = 20 cm = 0.2 m and $\theta = 30^{\circ}$. So, $Wg = 2 \times 10 \times 0.2 \times \cos(30^{\circ}) = 2 \, \text{J}$. Show more…
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