JEE Advanced · Physics · 6. Work Power Energy
A block of mass \(2 \mathrm{~kg}\) is free to move along the \(x\)-axis. It is at rest and from \(t\) \(=0\) onwards it is subjected to a time-dependent force \(F(t)\) in the \(x\) direction. The force \(F(t)\) varies with \(t\) as shown in the figure. The kinetic energy of the block after \(4.5 \mathrm{~s}\) is
- A \(4.50 \mathrm{~J}\)
- B \(7.50 \mathrm{~J}\)
- C \(5.06 \mathrm{~J}\)
- D \(14.06 \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(5.06 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Area under \(F\) - \(t\) graph \(=\) momentum
\(
\begin{aligned}
&=P=\sqrt{2 k m} \\
& \therefore \quad k=\frac{A^2}{2 m} \\
&(A=\text { net area of } F-t \text { graph) } \\
&=\frac{\left\{\left(\frac{4 \times 3}{2}\right)-\left(\frac{1.5 \times 2}{2}\right)\right\}^2}{2 \times 2}=5.0625 \mathrm{~J}
\end{aligned}
\)
\(\therefore\) The correct option is (c).
\(
\begin{aligned}
&=P=\sqrt{2 k m} \\
& \therefore \quad k=\frac{A^2}{2 m} \\
&(A=\text { net area of } F-t \text { graph) } \\
&=\frac{\left\{\left(\frac{4 \times 3}{2}\right)-\left(\frac{1.5 \times 2}{2}\right)\right\}^2}{2 \times 2}=5.0625 \mathrm{~J}
\end{aligned}
\)
\(\therefore\) The correct option is (c).
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