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JEE Physics 2023 Question with Solution

If the maximum load carried by an elevator is 1400 kg1400 \, \text{kg} (600 kg600 \, \text{kg} - Passengers + 800 kg800 \, \text{kg} - elevator), which is moving up with a uniform speed of 3 m s−13 \, \text{m s}^{-1} and the frictional force acting on it is 2000 N2000 \, \text{N}, then the maximum power used by the motor is _____ kW\text{kW} (g=10 m/s2g = 10 \, \text{m/s}^2).

Answer

Correct answer:48

Step-by-step solution

Standard Method

Given:

  • Mass of the elevator system = 1400 kg1400 \, \text{kg}
  • Velocity = 3 m/s3 \, \text{m/s}
  • Frictional force = 2000 N2000 \, \text{N}
  • Acceleration due to gravity = 10 m/s210 \, \text{m/s}^2

Find: Maximum power used by the motor.

Since the elevator is moving with uniform speed, the net force on it is zero. Therefore, the upward tension balances the weight and friction.

T=Mg+f=1400×10+2000=14000+2000=16000 NT = Mg + f = 1400 \times 10 + 2000 = 14000 + 2000 = 16000 \, \text{N}

Power used by the motor is:

P=F×V=T×V=16000×3=48000 W=48 kWP = F \times V = T \times V = 16000 \times 3 = 48000 \, \text{W} = 48 \, \text{kW}

Therefore, the maximum power used by the motor is 48 kW48 \, \text{kW}.

Force Balance and Power Relation

Given: The elevator moves upward with constant speed, so acceleration is zero.

Find: The motor power at maximum load.

For zero acceleration,

Upward force=downward forces\text{Upward force} = \text{downward forces}

So,

T=Mg+fT = Mg + f

Substituting the values,

T=1400×10+2000=16000 NT = 1400 \times 10 + 2000 = 16000 \, \text{N}

Now use the power formula,

P=TvP = TvP=16000×3=48000 WP = 16000 \times 3 = 48000 \, \text{W}

Converting into kilowatt,

48000 W=48 kW48000 \, \text{W} = 48 \, \text{kW}

Thus, the required numerical answer is 48.

Common mistakes

  • Taking the net force as non-zero because the elevator is moving upward. This is wrong because uniform speed means zero acceleration. Use force balance, not F=maF = ma with non-zero aa.

  • Ignoring the frictional force while calculating the tension. This is wrong because friction acts downward and the motor must also overcome it. Use T=Mg+fT = Mg + f.

  • Using weight alone in the power formula. This is wrong because the motor force is the tension, not just MgMg. First find the total opposing force, then use P=FvP = Fv.

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