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)
320 watts
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.
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
Power is the work done per unit of time. The formula for power is:
Power (P) = $\frac{\text{Work done (W)}}{\text{Time taken (t)}}$
We are given the following values:
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.
The power spent by the man is 320 watts.
Which one of the following is the value of 1 KWh of energy converted into joules?
Which of the following is correct regarding electric power?
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)
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.
Units of power is: