TS EAMCET · Physics · Work Power Energy
Under action of force, a \(2 \mathrm{~kg}\) body moves such that its position \(x\) as function of time \(t\) is given by \(x=\alpha t^2 / 2\), where \(x\) is in metre, \(t\) is in seconds and \(\alpha=1 \mathrm{~m} / \mathrm{s}^2\). The work done by the force in the first two seconds is
- A \(4 \mathrm{~J}\)
- B \(16 \mathrm{~J}\)
- C \(40 \mathrm{~J}\)
- D \(2 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(4 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Given, \(x=\frac{\alpha t^2}{2}\) As we know that, \(v=\frac{d x}{d t}=\frac{d}{d t}\left(\frac{\alpha t^2}{2}\right)=\alpha t\) \(\ldots(i)\) According to work-energy theorem Work done, \(W =\frac{1}{2} m v^2=\frac{1}{2} m(\alpha t)^2 \quad \text { [using Eq. (i)] }\)…
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