Search any question & find its solution
Question:
Answered & Verified by Expert
An object of mass $500 \mathrm{~g}$, initially at rest acted upon by a variable force whose $\mathrm{X}$ component varies with $\mathrm{X}$ in the manner shown. The velocities of the object a point $X=8 \mathrm{~m}$ and $X=12 \mathrm{~m}$, would be the respective values of (nearly)
Options:
Solution:
1529 Upvotes
Verified Answer
The correct answer is:
$23 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$
The area under the force displacement curve give the amount of work done.
From work-energy theorem,
$$
\begin{aligned}
& \mathrm{W}=\Delta \mathrm{KE} \\
& \therefore \text { At } \mathrm{x}=8 \mathrm{~m} \text {, } \\
& \mathrm{W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD } \\
& =20 \times 5+10 \times 3=130 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \Rightarrow \quad 130=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times \frac{500}{1000} \mathrm{v}^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{130}=22.8 \mathrm{~ms}^{-1} \approx 23 \mathrm{~ms}^{-1} \\
& \text { At } \quad \mathrm{x}=12 \mathrm{~m} \\
& \mathrm{~W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD + Area FGHIJ } \\
& + \text { Area KLMJ } \\
& \mathrm{W}=20 \times 5+10 \times 3+(-20 \times 2)+\left(\frac{1}{2} \times-5 \times 2\right) \\
& +10 \times 2 \quad[\because \text { Area FGHIJ }=\text { Area } \\
& \mathrm{FGIJ}+\text { Area } \mathrm{GHI}] \\
& =100+30-40-5+20=105 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \therefore \quad 105=\frac{1}{2} \times \frac{1}{2} \times v^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{105} \simeq 20.6 \mathrm{~ms}^{-1} \\
&
\end{aligned}
$$
From work-energy theorem,
$$
\begin{aligned}
& \mathrm{W}=\Delta \mathrm{KE} \\
& \therefore \text { At } \mathrm{x}=8 \mathrm{~m} \text {, } \\
& \mathrm{W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD } \\
& =20 \times 5+10 \times 3=130 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \Rightarrow \quad 130=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times \frac{500}{1000} \mathrm{v}^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{130}=22.8 \mathrm{~ms}^{-1} \approx 23 \mathrm{~ms}^{-1} \\
& \text { At } \quad \mathrm{x}=12 \mathrm{~m} \\
& \mathrm{~W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD + Area FGHIJ } \\
& + \text { Area KLMJ } \\
& \mathrm{W}=20 \times 5+10 \times 3+(-20 \times 2)+\left(\frac{1}{2} \times-5 \times 2\right) \\
& +10 \times 2 \quad[\because \text { Area FGHIJ }=\text { Area } \\
& \mathrm{FGIJ}+\text { Area } \mathrm{GHI}] \\
& =100+30-40-5+20=105 \mathrm{~J} \\
& \text { Using Eq. (i) } \\
& \therefore \quad 105=\frac{1}{2} \times \frac{1}{2} \times v^2 \\
& \Rightarrow \quad \mathrm{v}=2 \sqrt{105} \simeq 20.6 \mathrm{~ms}^{-1} \\
&
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.