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Question

A man of mass 80 kg climbs up 4 m high stairs in 10 sec. Find the power spent by the man. (Take g = 10 m/sec2)

The correct answer is

320 watts

Calculating Power Spent by the Man

This question asks us to find the power spent by a man while climbing stairs. Power is defined as the rate at which work is done. To solve this, we first need to calculate the work done by the man against gravity, and then divide it by the time taken.

Understanding Work Done

When the man climbs stairs, he is doing work against the force of gravity. The work done in lifting an object against gravity is calculated using the formula:

Work (W) = Force $\times$ Distance

In this case, the force is the weight of the man (mass $\times$ acceleration due to gravity), and the distance is the height of the stairs.

Weight of the man (Force) = mass (m) $\times$ gravity (g)

So, Work done (W) = m $\times$ g $\times$ h

Understanding Power

Power is the work done per unit of time. The formula for power is:

Power (P) = $\frac{\text{Work done (W)}}{\text{Time taken (t)}}$

Applying the Formulas

We are given the following values:

  • Mass of the man (m) = 80 kg
  • Height of the stairs (h) = 4 m
  • Time taken (t) = 10 sec
  • Acceleration due to gravity (g) = 10 m/sec$^2$

First, let's calculate the work done:

W = m $\times$ g $\times$ h

W = 80 kg $\times$ 10 m/sec$^2$ $\times$ 4 m

W = 800 N $\times$ 4 m

W = 3200 Joules

Now, let's calculate the power spent:

P = $\frac{\text{W}}{\text{t}}$

P = $\frac{3200 \text{ Joules}}{10 \text{ sec}}$

P = 320 Joules/sec

Since 1 Joule/sec = 1 Watt, the power is 320 Watts.

Summary of Calculation

  • Work done (W) = mgh = (80 kg)(10 m/s$^2$)(4 m) = 3200 J
  • Power (P) = W/t = $\frac{3200 \text{ J}}{10 \text{ s}} = 320 \text{ W}$

The power spent by the man is 320 watts.

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Important Questions from Power

  1. Which one of the following is the value of 1 KWh of energy converted into joules?

  2. Which of the following is correct regarding electric power?

  3. An elevator weighing 400 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used? (g = 9.8 m/s 2and there is no frictional loss)

  4. An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s2 and there is no frictional loss.

  5. Units of power is:

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