Question

A block of mass $5\text{kg}$ which is at rest can slide along the smooth track. A man pulls the block through a rope having tension $10\text{N}$ which makes an angle of ${60}^{\circ }$ with horizontal. If the block moves $5\text{m}$ along horizontal, the amount of work done by tension is

• A. $25\text{J}$
• B. $50\text{J}$
• C. $12.5\text{J}$
• D. $-25\text{J}$

Answer 1

The correct option is A $25\text{J}$

Horizontal component of tension $\left(T_x\right)=T\cos {60}^{\circ }$
$=\frac{T}{2}=\frac{10}{2}=5\text{N}$
Vertical component of tension $\left(T_y\right)=T\sin {60}^{\circ }$
$=\frac{T\sqrt{3}}{2}=\frac{10\sqrt{3}}{2}=5\sqrt{3}\text{N}$
From FBD of the block:
Using $\sum F_y=0$
$R+5\sqrt{3}=50$

As the block moves only in horizontal direction, amount of work done by vertical component of tension is equal to zero. ($\because T_y\perp s$)
Since, both $T_x$ and $s$ are in the same direction (parallel)
Workdone by tension $\left(W\right)=T_{x}s=5\times 5=25\text{J}$

Hence option A is the correct answer