COMEDK · Physics · 8. Rotational Motion
The diagram shows a barrel of weight \(1.0 \times 10^{3} \mathrm{~N}\) on a frictionless slope inclined at \(30^{\circ}\) to the horizontal.
The force is parallel to the slope. What is the work done in moving the barrel a distance of \(5.0 \mathrm{~m}\) up the slope?
- A \(2.5 \times 10^{3} \mathrm{~J}\)
- B \(4.3 \times 10^{3} \mathrm{~J}\)
- C \(5.0 \times 10^{3} \mathrm{~J}\)
- D \(1.0 \times 10^{3} \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(2.5 \times 10^{3} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below Work done in moving the barrel on the frictionless slope is equal to change in potential energy. i.e. \(W=m g\left(h_{1}-h_{2}\right)\) Here, \(\quad m g=1 \times 10^{3} \mathrm{~N}\) From figure, \(\sin 30^{\circ}=\frac{h_{1}-h_{2}}{5}\)…
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