Question

A man (mass=68 kg) on a parachute is falling at terminal velocity (v=59 m/s)

At which rate do the internal energy of the man and of the air around him increase?

Answer 1

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

$E=mgh$

Differentiating equation (1) wrt t as :

$\dfrac{dE}{dt}=mg\dfrac{dh}{dt}$

Since, $v=\dfrac{dh}{dt}=59\ m/s$

$\dfrac{dE}{dt}=68\times 9.8\times 59$

$\dfrac{dE}{dt}=39317.6\ J/s$

So, the internal energy of the man and the air around him is increasing at the rate of 39317.6 J/s. Hence, this is the required solution.