JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A force \(\mathrm{F}=\alpha+\beta \mathrm{x}^2\) acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant \(\alpha=1 \mathrm{~N}\) then \(\beta\) will be ________.
- A \(15 \mathrm{~N} / \mathrm{m}^2\)
- B \(12 \mathrm{~N} / \mathrm{m}^2\)
- C \(8 \mathrm{~N} / \mathrm{m}^2\)
- D \(10 \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(B) \(12 \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
\(F=\alpha+\beta x^2\) Work done \(\int d w=\int F \cdot d x\)…
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