TS EAMCET · Physics · Laws of Motion
A block of mass 0.5 kg is at rest on a horizontal table. The coefficient of kinetic friction between the table and the block is 0.2 . If a horizontal force of 5 N is applied on the block, the kinetic energy of the block in a time of 4 s is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) ),
- A 64 J
- B 128 J
- C 256 J
- D 512 J
Answer & Solution
Correct Answer
(C) 256 J
Step-by-step Solution
Detailed explanation
Here force applied \(\mathrm{F}_{\text {app }}=5 \mathrm{~N}, \mathrm{~m}=0.5 \mathrm{~kg}, \mu_{\mathrm{f}}=0.2\) \(\mathrm{a}=\frac{\mathbf{F}_{\mathrm{app}}-\mathbf{f}_{\mathrm{k}}}{\mathrm{~m}}\) or,…
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