Work, Energy, and Simple Machines: CBSE Class 9 Physics practice questions

B. $200\ \text{J}$

Work done $= F \times s$ with $s$ in metres: $400\ \text{cm} = 4\ \text{m}$, so $W = 50\ \text{N} \times 4\ \text{m} = 200\ \text{J}$. Using 400 straight from centimetres gives the trap answer $20{,}000\ \text{J}$. $12.5\ \text{J}$ comes from dividing force by distance, and $54\ \text{J}$ from adding them.

D. Negative, because the dumbbell moves opposite to her upward force

While lowering it slowly, she still pushes upwards (about equal to its weight), but the dumbbell moves downwards. The force and displacement are opposite, so her work on it is negative. While raising it, force and displacement are in the same direction, so that work is positive. The work is not zero, because the dumbbell does move.

A. $60\ \text{m s}^{-1}$

Work by the wire $= -2{,}00{,}000\ \text{N} \times 90\ \text{m} = -1{,}80{,}00{,}000\ \text{J}$. By the work-energy theorem this equals $0 - \tfrac{1}{2}(10{,}000)v^2$, so $v^2 = \dfrac{2 \times 1.8 \times 10^7}{10{,}000} = 3600$ and $v = 60\ \text{m s}^{-1}$ (216 km/h). Forgetting the $\tfrac{1}{2}$ gives $v^2 = 1800$, about $42\ \text{m s}^{-1}$. Forgetting the square root gives 3600, and making both slips gives 1800.