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NVAEasyJEE 2023Motion in a Straight Line

JEE Physics 2023 Question with Solution

A lift of mass M=500 kgM = 500 \, \text{kg} is descending with a speed of 2 m/s2 \, \text{m/s}. Its supporting cable begins to slip, thus allowing it to fall with a constant acceleration of 2 m/s22 \, \text{m/s}^2. The kinetic energy of the lift at the end of fall through a distance of 6 m6 \, \text{m} will be ..... kJ.

Answer

Correct answer:7

Step-by-step solution

Standard Method

Given: mass of lift M=500 kgM = 500 \, \text{kg}, initial speed u=2 m/su = 2 \, \text{m/s} downward, constant acceleration a=2 m/s2a = 2 \, \text{m/s}^2 downward, distance fallen s=6 ms = 6 \, \text{m}.

Find: kinetic energy of the lift after falling 6 m6 \, \text{m}.

Use the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

Substituting the values:

v2=22+2(2)(6)v^2 = 2^2 + 2(2)(6) v2=4+24=28v^2 = 4 + 24 = 28

Now kinetic energy is

KE=12mv2KE = \frac{1}{2}mv^2

Substituting:

KE=12(500)(28)KE = \frac{1}{2}(500)(28) KE=7000 J=7 kJKE = 7000 \, \text{J} = 7 \, \text{kJ}

Therefore, the kinetic energy of the lift is 7 kJ7 \, \text{kJ}.

Velocity First, Then Energy

Given: u=2 m/su = 2 \, \text{m/s}, a=2 m/s2a = 2 \, \text{m/s}^2, s=6 ms = 6 \, \text{m}, m=500 kgm = 500 \, \text{kg}.

Find: final kinetic energy.

First calculate the final speed from kinematics:

v2=u2+2as=22+2×2×6=28v^2 = u^2 + 2as = 2^2 + 2 \times 2 \times 6 = 28

So,

v=28=5.29 m/sv = \sqrt{28} = 5.29 \, \text{m/s}

Now apply the kinetic energy formula:

KE=12mv2KE = \frac{1}{2}mv^2 KE=12×500×28=7000 JKE = \frac{1}{2} \times 500 \times 28 = 7000 \, \text{J}

Converting to kilojoule:

7000 J=7 kJ7000 \, \text{J} = 7 \, \text{kJ}

Thus, the required numerical answer is 7.

Common mistakes

  • Using vv instead of v2v^2 in the kinetic energy substitution after already finding v2=28v^2 = 28. This is wrong because KE=12mv2KE = \frac{1}{2}mv^2 directly uses v2v^2. Substitute 2828 directly, or first find v=28v = \sqrt{28} and then square it consistently.

  • Taking the initial speed as negative without keeping the sign convention consistent. This is wrong because both motion and acceleration are downward here, so a consistent positive downward convention makes u=2u = 2 and a=2a = 2. Choose one direction and use it throughout.

  • Forgetting to convert joules to kilojoules at the end. This is wrong because 7000 J7000 \, \text{J} must be written as 7 kJ7 \, \text{kJ}. Divide by 10001000 before writing the final answer.

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