A block of mass $2 \mathrm{~kg}$ is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at $30^{\circ}$. Then the work done by the tension in the rope in pulling it a distance $4 \mathrm{~m}$ along the inclined plane in joule is $\left(\right.$ acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

• Correct answer
  A

  $40$
• B

  $20$
• C

  $68$
• D

  $136$

Solution

As $V=$ constant $\begin{aligned} & \Rightarrow \mathrm{T}=\mathrm{mg} \sin \theta \\ & \Rightarrow \mathrm{T}=2 \times 10 \times \sin 30^{\circ} \\ & \Rightarrow \mathrm{T}=10 \mathrm{~N}\end{aligned}$ So, work done $=\mathrm{T} \times \mathrm{x}$ $=10 \times 4=40 \mathrm{~N}$

Hence, the correct answer is option: 'A'.