Question

A hand-driven tire pump has a piston with a 2.50-cm diameter and a maximum stroke of 30.0 cm. (a) How much work do you do in one stroke if the average pressure is 2.40 ×10^{5} 5 N/m^2 2 ? (b) What average force do you exert on the piston, neglecting friction and gravitational force?

Answer 1

(a) you do a work on the piston: 35.34 (Joules) and (b) Average Force that you exert on the piston is: 117.81 (N)

Step-by-step explanation:

We need to know that the work do it for a piston or boundary work of closed system is:
`W=P*DeltaV` where W is the work, P is the pressure and delta_V is the change in the volume. Now we need to remember that the volume for a piston is the same as:
`Volumen=(\pi *diameter^(2) )/(4) *L` where L is the maximum stroke or length, so the piston volume is:
`Volume=(\pi*0.025^(2))/(4)*0.3=0.000147(metres^(3) )` therefore, piston's work is:
`W=2.4*10^(5) *0.000147=35.34(Jolues)`. Then we are going to calculate the average force so we need to know that
`P=(F)/(A)` where P is the pressure, F is the force and A the area in this case the piston´s area, so
`A=(\pi*diameter^(2))/(4)= (\pi*0.025^(2) )/(4) =0.00049(metres^(2))` and replacing values we get:
`F=P*A=2.4*10^(5)*0.00049=117.81 (N)`