Answer to Question #143469 in Mechanics | Relativity for tejanshi tejanshi

Question

 A 60 kg student runs up a 300m long, 8.5° ramp at a constant speed of 2.5 m/s on a 25°C day. The student has a surface area of 1.50 m2, 34°C skin temperature, a skin emissivity of 0.90 and muscles that are 25% efficient. The surface transfer coefficient is 8 watts per square metre-degree. If resistance forces on the student amount to 54 N, then find a) the force required to run up the ramp b) the rate at which the student does work c) the rate at which heat energy is generated by the student (joules per second) d) the rate of heat radiation by the skin e) the rate of convective cooling f) the rate of sweating (kg/s) required to maintain constant body temperature given the rate of exercise, radiative emission, and convective heat losses.

Answer 1

A) since the work, without resistance and loss, is equal to the change in kinetic energy, $\Delta E_c=A$ , hence $\frac{m\times v^2 }{2} =F\times S$

$F=\frac{m \times v^2}{2\times S}=\frac{60 \times 2.5^2}{2\times 300}=0.625$ N.

Strength with resistance and muscle loss $F_{total}=\frac{0.625+54}{0.25}=218.5$ N.

B) since the power is $P=F\times v$ , then $P=54.625\times 2.5=136.5625$ J/s.

c) Similar to the previous paragraph $P=(218.5-54.625)\times 2.5=409.6875$ J/s.

d) The rate of thermal radiation by the skin is

$p=(t_b-t_a)\times \times x \times S$ $=(34 -25)\times 0.9 \times 8 \times 1.5=97.2$ J/s.

e) the rate of convective cooling is determined by the formula where $\theta=4.06 \frac{Vt}{m^2\times C\degree}$ is the constant $Q=\theta\times S (t_b-t_a)=4.06\times 1.5 (34-25)=54.81$ J/s.

f) the rate of sweating is equal to the difference between the energy produced and the energy released into the medium: $V=409.6875-(54.81+97.2)=257.6775$ J/s.