An object of mass 500 ; g, initially at rest, | vedclass

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(C) The work done by the force is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^2$. The mass $m = 500 \; g = 0.5 \; kg$.$1$.

Vedclass · Accepted Answer

$23 \; m/s$ and $20.6 \; m/s$

Vedclass · Answer

(C) The work done by the force is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^2$. The mass $m = 500 \; g = 0.5 \; kg$.$1$. Work done up to $X = 8 \; m$:$W_8 = (20 \; N 	imes 5 \; m) + (10 \; N 	imes 3 \; m) = 100 + 30 = 130 \; J$.Using $W = \frac{1}{2} m v^2$: $130 = \frac{1}{2} (0.5) v_8^2 \Rightarrow v_8^2 = 520 \Rightarrow v_8 = \sqrt{520} \approx 22.8 \; m/s \approx 23 \; m/s$.$2$. Work done up to $X = 12 \; m$:$W_{12} = W_8 + (	ext{Area from } 8 \; m 	ext{ to } 10 \; m) + (	ext{Area from } 10 \; m 	ext{ to } 12 \; m)$.Area from $8 \; m$ to $10 \; m = (2 \; m) 	imes (-25 \; N) = -50 \; J$.Area from $10 \; m$ to $12 \; m = (2 \; m) 	imes (10 \; N) = 20 \; J$.$W_{12} = 130 - 50 + 20 = 100 \; J$.Using $W = \frac{1}{2} m v^2$: $100 = \frac{1}{2} (0.5) v_{12}^2 \Rightarrow v_{12}^2 = 400 \Rightarrow v_{12} = 20 \; m/s \approx 20.6 \; m/s$ (closest option).Thus, the velocities are approximately $23 \; m/s$ and $20.6 \; m/s$.

An object of mass $500 \; g$, initially at rest, is acted upon by a variable force whose $X$-component varies with $X$ as shown in the graph. The velocities of the object at $X = 8 \; m$ and $X = 12 \; m$ are, respectively:

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