JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A body of mass ' \({m}\) ' dropped from a height ' \({h}\) ' reaches the ground with a speed of \(0.8 \sqrt{{gh}}\). The value of workdone by the air-friction is \(.....\,{mgh}\)
- A \(0.68\)
- B \(1\)
- C \(1.64\)
- D \(0.64\)
Answer & Solution
Correct Answer
(A) \(0.68\)
Step-by-step Solution
Detailed explanation
Work done \(=\) Change in kinetic energy \({W}_{{mg}}+{W}_{{air}-\text { fiction }}=\frac{1}{2} {m}(.8 \sqrt{{gh}})^{2}-\frac{1}{2} {m}(0)^{2}\) \({W}_{\text {air - fiction }}=\frac{.64}{2} {mgh}-{mgh}=0.68 {mgh}\)
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