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Work Power and Energy question

2025 · 28 Jan · Shift 2 · Q63
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  5. /2025 · 28 Jan · Shift 2 · Q63

Work Power and Energy question

2025 · 28 Jan · Shift 2 · Q63

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A body of mass 4 kg is placed on a plane at a point PPP having coordinate (3,4)m(3,4) \mathrm{m}(3,4)m. Under the action of force F→=(2i^+3j^)N\overrightarrow{\mathrm{F}}=(2 \hat{i}+3 \hat{j}) \mathrm{N}F=(2i^+3j^​)N, it moves to a new point Q having coordinates (6,10)m(6,10) \mathrm{m}(6,10)m in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :
  1. A
    4 : 3
  2. B
    13 : 6
  3. C
    1 : 2
  4. D
    6 : 13
View written solutionFree

Correct answer: 12 : 25

  1. Given data
  • Mass of body: m=4 kgm=4\,\text{kg}m=4kg
  • Initial position: P(3,4)P(3,4)P(3,4)
  • Final position after 4 4\,4s: Q(6,10)Q(6,10)Q(6,10)
  • Force: F⃗=(2i^+3j^) N\vec F=(2\hat i+3\hat j)\,\text{N}F=(2i^+3j^​)N
  • Time taken: t=4 t=4\,t=4s

We need the ratio: average power:instantaneous power at t=4 s\text{average power} : \text{instantaneous power at } t=4\text{ s}average power:instantaneous power at t=4 s


  1. Displacement from PPP to QQQ

Δr⃗=(6−3)i^+(10−4)j^=3i^+6j^\Delta \vec r=(6-3)\hat i+(10-4)\hat j=3\hat i+6\hat jΔr=(6−3)i^+(10−4)j^​=3i^+6j^​


  1. Average power

Average power is Pavg=work donetimeP_{\text{avg}}=\frac{\text{work done}}{\text{time}}Pavg​=timework done​

Work done by constant force: W=F⃗⋅Δr⃗W=\vec F\cdot \Delta \vec rW=F⋅Δr

So, W=(2i^+3j^)⋅(3i^+6j^)=2⋅3+3⋅6=6+18=24 JW=(2\hat i+3\hat j)\cdot(3\hat i+6\hat j)=2\cdot 3+3\cdot 6=6+18=24\,\text{J}W=(2i^+3j^​)⋅(3i^+6j^​)=2⋅3+3⋅6=6+18=24J

Hence, Pavg=244=6 WP_{\text{avg}}=\frac{24}{4}=6\,\text{W}Pavg​=424​=6W


  1. Find acceleration

Using Newton's second law, a⃗=F⃗m\vec a=\frac{\vec F}{m}a=mF​

Thus, a⃗=14(2i^+3j^)=(12i^+34j^) m/s2\vec a=\frac{1}{4}(2\hat i+3\hat j)=\left(\frac12\hat i+\frac34\hat j\right)\,\text{m/s}^2a=41​(2i^+3j^​)=(21​i^+43​j^​)m/s2


  1. Find initial velocity

Use Δr⃗=u⃗t+12a⃗t2\Delta \vec r=\vec u t+\frac12\vec a t^2Δr=ut+21​at2

Given t=4t=4t=4 s, 3i^+6j^=4u⃗+12a⃗(16)3\hat i+6\hat j=4\vec u+\frac12\vec a(16)3i^+6j^​=4u+21​a(16) 3i^+6j^=4u⃗+8a⃗3\hat i+6\hat j=4\vec u+8\vec a3i^+6j^​=4u+8a

Now, 8a⃗=8(12i^+34j^)=4i^+6j^8\vec a=8\left(\frac12\hat i+\frac34\hat j\right)=4\hat i+6\hat j8a=8(21​i^+43​j^​)=4i^+6j^​

So, 3i^+6j^=4u⃗+4i^+6j^3\hat i+6\hat j=4\vec u+4\hat i+6\hat j3i^+6j^​=4u+4i^+6j^​

Therefore, 4u⃗=−i^4\vec u=-\hat i4u=−i^ u⃗=−14i^\vec u=-\frac14\hat iu=−41​i^


  1. Velocity at the end of 4 s

v⃗=u⃗+a⃗t\vec v=\vec u+\vec a tv=u+at

So, v⃗=−14i^+4(12i^+34j^)\vec v=-\frac14\hat i+4\left(\frac12\hat i+\frac34\hat j\right)v=−41​i^+4(21​i^+43​j^​) v⃗=−14i^+2i^+3j^\vec v=-\frac14\hat i+2\hat i+3\hat jv=−41​i^+2i^+3j^​ v⃗=74i^+3j^\vec v=\frac74\hat i+3\hat jv=47​i^+3j^​


  1. Instantaneous power at t=4t=4t=4 s

Instantaneous power is Pinst=F⃗⋅v⃗P_{\text{inst}}=\vec F\cdot \vec vPinst​=F⋅v

Thus, Pinst=(2i^+3j^)⋅(74i^+3j^)P_{\text{inst}}=(2\hat i+3\hat j)\cdot\left(\frac74\hat i+3\hat j\right)Pinst​=(2i^+3j^​)⋅(47​i^+3j^​) Pinst=2⋅74+3⋅3P_{\text{inst}}=2\cdot\frac74+3\cdot 3Pinst​=2⋅47​+3⋅3 Pinst=72+9=252 WP_{\text{inst}}=\frac72+9=\frac{25}{2}\,\text{W}Pinst​=27​+9=225​W


  1. Required ratio

Pavg:Pinst=6:252=12:25P_{\text{avg}}:P_{\text{inst}}=6:\frac{25}{2}=12:25Pavg​:Pinst​=6:225​=12:25

So the required ratio is 12:25\boxed{12:25}12:25​


  1. Comparison with options

Given options are:

  • A: 4:34:34:3
  • B: 13:613:613:6
  • C: 1:21:21:2
  • D: 6:136:136:13

Our derived ratio 12:2512:2512:25 does not match any option.


  1. Check stored answer

Stored correct answer: D (6:136:136:13)

But our calculated answer is 12:2512:2512:25, so I do not agree with the stored answer.

A likely reason is that the question/options may contain a typo or some intended assumption is missing. Under the standard interpretation with constant force and motion under that force, the ratio is definitively 12:2512:2512:25.

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