JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A block of mass \(1 \mathrm{~kg}\) is pushed up a surface inclined to horizontal at an angle of \(60^{\circ}\) by a force of \(10 \mathrm{~N}\) parallel to the inclined surface as shown in figure. When the block is pushed up by \(10 \mathrm{~m}\) along inclined surface, the work done against frictional force is _______. \(\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]\)
- A \(5 \sqrt{3} \mathrm{~J}\)
- B \(5 \mathrm{~J}\)
- C \(5 \times 10^3 \mathrm{~J}\)
- D \(10 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Work done against frictional force \(=\mu \mathrm{N} \times 10\) \(=0.1 \times 5 \times 10=5 \mathrm{~J}\)
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