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MCQEasyJEE 2023Newton's Second Law & Force

JEE Physics 2023 Question with Solution

Three forces F1=10 N,F2=8 N,F3=6 NF_1 = 10 \, N, F_2 = 8 \, N, F_3 = 6 \, N are acting on a particle of mass 5 kg5 \, \text{kg}. The forces F2F_2 and F3F_3 are applied perpendicular so that particle remains at rest. If the force F1F_1 is removed, then the acceleration of the particle is:

  • A

    2 ms−22 \, \text{ms}^{-2}.

  • B

    0.5 ms−20.5 \, \text{ms}^{-2}.

  • C

    4.8 ms−24.8 \, \text{ms}^{-2}.

  • D

    7 ms−27 \, \text{ms}^{-2}.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Three forces F1=10 NF_1 = 10 \, N, F2=8 NF_2 = 8 \, N and F3=6 NF_3 = 6 \, N act on a particle of mass 5 kg5 \, \text{kg}. Initially the particle remains at rest, so the body is in equilibrium.

Find: The acceleration when F1F_1 is removed.

Since the body is in equilibrium, the net force acting on it is zero.

F1+F2+F3=0F_1 + F_2 + F_3 = 0

If F1F_1 is removed, the particle experiences acceleration due to the remaining perpendicular forces F2F_2 and F3F_3.

The magnitude of the resultant force is

∣Fnet∣=F22+F32=62+82=10 N|F_{\text{net}}| = \sqrt{F_2^2 + F_3^2} = \sqrt{6^2 + 8^2} = 10 \, N

The solution also gives the direction through

tan⁡θ=F3F2=86\tan \theta = \frac{F_3}{F_2} = \frac{8}{6}θ=tan⁡−1(86)=53.13∘\theta = \tan^{-1}\left(\frac{8}{6}\right) = 53.13^{\circ}

Now use Newton's second law:

Fnet=maF_{\text{net}} = maa=Fnetm=105=2 ms2a = \frac{F_{\text{net}}}{m} = \frac{10}{5} = 2 \, \frac{m}{s^2}

Therefore, the acceleration of the particle is 2 m/s22 \, \text{m/s}^2. The correct option is A.

Equilibrium to resultant force

Given: The particle is initially at rest under three forces, so the three forces balance each other.

Find: Acceleration after removing F1F_1.

Because the initial net force is zero, F1F_1 must be equal and opposite to the resultant of F2F_2 and F3F_3. After removing F1F_1, only that resultant remains.

Since F2F_2 and F3F_3 are perpendicular,

R=82+62=64+36=100=10 NR = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, N

Then

a=Rm=105=2 m/s2a = \frac{R}{m} = \frac{10}{5} = 2 \, \text{m/s}^2

Therefore, the particle accelerates with 2 m/s22 \, \text{m/s}^2, so the correct option is A.

Common mistakes

  • Treating F2F_2 and F3F_3 as if they act in the same line and adding or subtracting them directly is wrong because they are perpendicular. Use the resultant magnitude F22+F32\sqrt{F_2^2 + F_3^2} instead.

  • Using all three forces after F1F_1 is removed is incorrect. Once F1F_1 is removed, only F2F_2 and F3F_3 contribute to the new net force.

  • Forgetting that the body was initially in equilibrium can hide the key idea. The initial rest condition means the resultant of F2F_2 and F3F_3 was exactly balanced by F1F_1.

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