The minimum power required to lift a mass of 50 kg up to a vertical distance of 8 m in 10 s is: (Take g = 10 m/s²)

Correct — A, 400 W. Lifting a body at a steady rate means applying an upward force equal to its weight, mg, which here is 50 kilograms times 10 metres per second squared, or 500 newtons. Moving that force through 8 metres does 500 times 8 equals 4,000 joules of work — the same figure you get straight from mgh. Power is the rate of doing work, so dividing by the 10 seconds gives 400 watts. The word 'minimum' in the stem is doing real work: it tells you to ignore friction and any kinetic energy left in the load at the top, so that every joule supplied ends up as gravitational potential energy. Any real machine would need more than 400 watts; 400 is the floor.

• (b)40 W — A factor of ten low. It comes from dividing 4,000 joules by 100 seconds, or from dropping the g when computing the weight.
• (c)50 W — This is the mass in kilograms carried across as a power in watts. Mass alone is not enough — the weight, the height and the time all enter.
• (d)500 W — This is the weight of the load, 50 x 10 = 500 newtons, mistaken for the power. It still has to be multiplied by 8 metres and divided by 10 seconds.

Work is done when a force moves its point of application through a distance in the direction of the force, and one joule is the work done when a force of one newton moves an object one metre. Power is the rate at which work is done, measured in watts, one watt being one joule per second. Against gravity, the work needed to raise a mass m through a height h is mgh, and it does not depend on the path taken — only on the vertical rise — because gravity is a conservative force.

Numerical items of this shape are almost always testing unit discipline rather than physics. Two of the wrong options here are quantities computed correctly along the way and then mislabelled: 500 is the weight in newtons, and 50 is the mass in kilograms. A useful habit is to track units through the calculation rather than only the numbers — kilogram times metre per second squared gives newton, newton times metre gives joule, joule per second gives watt — and to notice that only the final line carries watts. The other habit is to read the qualifier: 'minimum' power, 'average' power and 'power delivered by the motor' are three different questions, and the last would require an efficiency figure.

• Work done against gravity in raising a mass m through a height h is mgh, independent of the path taken.
• Power is work divided by time; one watt is one joule per second.
• Here the work is 50 x 10 x 8 = 4,000 J and the power is 4,000 / 10 = 400 W.
• The weight of the load is mg = 500 N, which is a force in newtons and not a power in watts.
• 'Minimum' power assumes no friction and no leftover kinetic energy, so all the work supplied becomes potential energy.

• Reporting the weight in newtons as the answer in watts.
• Forgetting to divide by the time, which turns a power question into an energy question.
• Using the given g of 10 m/s squared inconsistently, or slipping in 9.8 partway through.

As a short numerical on power, work or energy with the value of g supplied, or as a unit-conversion question involving the kilowatt hour.

• practice — not a real PYQ

  A pump raises 100 kg of water through a vertical height of 10 m in 20 s. The minimum power of the pump is: (Take g = 10 m/s²)

  • (a)100 W
  • (b)500 W
  • (c)1000 W
  • (d)2000 W

  Answer(b) 500 W — the work done is mgh = 100 x 10 x 10 = 10,000 J, and 10,000 J divided by 20 s is 500 W.

• practice — not a real PYQ

  One watt is equivalent to which one of the following?

  • (a)One newton per second
  • (b)One joule per second
  • (c)One joule metre
  • (d)One newton metre

  Answer(b) One joule per second — power is the rate of doing work; a newton metre is a joule, which is energy, not power.