TS EAMCET · Physics · Work Power Energy
The block starts from rest as shown in the figure. Find the work done by force of \(10 \mathrm{I}\) and friction in the time 0 to \(4 \mathrm{~s}\). [Take, \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A \(240 \mathrm{~J},-96 \mathrm{~J}\)
- B \(250 \mathrm{~J}, 96 \mathrm{~J}\)
- C \(240 \mathrm{~J}, 96 \mathrm{~J}\)
- D \(250 \mathrm{~J},-96 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(240 \mathrm{~J},-96 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Here, applied force, \(F=10 \mathrm{~N}\) Friction force, \(f=\mu N=0.2 \times m g=0.2 \times 2 \times 10=4 \mathrm{~N}\) Net force, \(F_{\text {net }}=F-f=10-4=6 \mathrm{~N}\) \(\therefore\) Acceleration, \(a=\frac{F_{\text {net }}}{m}=\frac{6}{2}=3 \mathrm{~ms}^{-2}\) Distance…
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