Question

A block of mass $10\text{kg}$ pulled by force $100\text{N}$. It covers a distance $500\text{m}$ in $10s$. From initial point this motion is observed by two observers $A$ and $B$ as shown in fig.



Find out work done by force $F$ in 10 seconds w.r.t. $A$ and $B$ respectively.

• A. $5000\text{J}$ and $5000\text{J}$
• B. $5000\text{J}$ and $0$
• C. $4000\text{J}$ and $0$
• D. $0$ and $6000\text{J}$

Answer 1

The correct option is C $4000\text{J}$ and $0$


Distance covered by block $=S$
Distance covered by $A=S_A$
$S_A=u_A\times t=10\times 10=100\text{m}$
Work done with respect to $A$ is
$(W_F{)}_{\text{w.r.t}A}=F[S-S_A]=100[500-100]$
$(W_F{)}_{\text{w.r.t}A}=4000\text{J}$
Distance covered by $B=S_B$
Now,
By applying $2^{nd}$ law of motion,
we get
$S_B=u_{B}t+\frac{1}{2}{a}_B{t}^2$
$S_B=0+\frac{1}{2}\times 10\times {10}^2$
$S_B=500\text{m}$
Workdone with respect to $B$ is
$(W_F{)}_{w.r.tB}=F[S-S_B]=100(500-500)$
$(W_F{)}_{w.r.tB}=0$

Answer $4000\text{J},0$