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NEET 2019
Exam
NEET
Year
2019
Subject
Physics
Chapter
Work, Power and Energy

NEET 2019 · Physics · Work, Power and Energy

An object of mass 500 ~g , initially at rest acted upon by a...

An object of mass 500 g500 \mathrm{~g}, initially at rest acted upon by a variable force whose X\mathrm{X} component varies with X\mathrm{X} in the manner shown. The velocities of the object a point X=8 mX=8 \mathrm{~m} and X=12 mX=12 \mathrm{~m}, would be the respective values of (nearly) An object of mass 500 ~g , initially at rest acted upon by a variable force whose X component varies with X in the

Options

  1. A18 m/s18 \mathrm{~m} / \mathrm{s} and 24.4 m/s24.4 \mathrm{~m} / \mathrm{s}
  2. B23 m/s23 \mathrm{~m} / \mathrm{s} and 24.4 m/s24.4 \mathrm{~m} / \mathrm{s}
  3. C23 m/s23 \mathrm{~m} / \mathrm{s} and 20.6 m/s20.6 \mathrm{~m} / \mathrm{s}
  4. D18 m/s18 \mathrm{~m} / \mathrm{s} and 20.6 m/s20.6 \mathrm{~m} / \mathrm{s}

Correct answer

C. 23 m/s23 \mathrm{~m} / \mathrm{s} and 20.6 m/s20.6 \mathrm{~m} / \mathrm{s}

Step-by-step solution

The area under the force displacement curve give the amount of work done. An object of mass 500 ~g , initially at rest acted upon by a variable force whose X component varies with X in the From work-energy theorem, W=ΔKE∴ At x=8 m, W= Area ABDO+ Area CEFD =20×5+10×3=130 J Using Eq. (i) ⇒130=12mv2=12×5001000v2⇒v=2130=22.8 ms−1≈23 ms−1 At x=12 m W= Area ABDO+ Area CEFD + Area FGHIJ + Area KLMJ W=20×5+10×3+(−20×2)+(12×−5×2)+10×2[∵ Area FGHIJ = Area FGIJ+ Area GHI]=100+30−40−5+20=105 J Using Eq. (i) ∴105=12×12\timesv2⇒v=2105≃20.6 ms−1\begin{aligned}& \mathrm{W}=\Delta \mathrm{KE}\\ & \therefore \text{ At } \mathrm{x}=8 \mathrm{~m} \text{, }\\ & \mathrm{W}=\text{ Area } \mathrm{ABDO}+\text{ Area CEFD }\\ & =20 \times5+10 \times3=130 \mathrm{~J}\\ & \text{ Using Eq. (i) }\\ & \Rightarrow \quad 130=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times \frac{500}{1000} \mathrm{v}^2 \\ & \Rightarrow \quad \mathrm{v}=2 \sqrt{130}=22.8 \mathrm{~ms}^{-1} \approx23 \mathrm{~ms}^{-1}\\ & \text{ At } \quad \mathrm{x}=12 \mathrm{~m}\\ & \mathrm{~W}=\text{ Area } \mathrm{ABDO}+\text{ Area CEFD + Area FGHIJ }\\ & + \text{ Area KLMJ }\\ & \mathrm{W}=20 \times5+10 \times₃⁺(-20 \times2)+\left(\frac{1}{2} \times-5 \times2\right) \\ & +10 \times2 \quad[\because \text{ Area FGHIJ }=\text{ Area }\\ & \mathrm{FGIJ}+\text{ Area } \mathrm{GHI}]\\ & =100+30-40-5+20=105 \mathrm{~J}\\ & \text{ Using Eq. (i) }\\ & \therefore \quad 105=\frac{1}{2} \times \frac{1}{2} \timesv^2 \\ & \Rightarrow \quad \mathrm{v}=2 \sqrt{105} \simeq20.6 \mathrm{~ms}^{-1}\\ & \end{aligned}
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