116k views
3 votes
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 gauge pressure is 2.40×105N/m2 (about 35 psi)? (b) What average force do you exert on the piston, neglecting friction and gravitational force?

a) 13.2 J, 50.2 N
b) 26.4 J, 100.4 N
c) 39.6 J, 150.6 N
d) 52.8 J, 200.8 N

User Adiog
by
7.6k points

1 Answer

4 votes

Final answer:

During one stroke of the tire pump, 35.34 Joules of work is done, achieved by exerting an average force of 117.81 Newtons on the piston.

Step-by-step explanation:

A hand-driven tire pump with a 2.50-cm diameter piston and a maximum stroke of 30.0 cm is used to inflate a tire. To determine the work done and the average force exerted on the piston during one stroke, we'll perform the following calculations:

First, we calculate the area of the piston using the formula πr², where r is the radius in meters.

Next, we use the given average gauge pressure to find the average force exerted by the pump using the formula Force = Pressure × Area.

Finally, we calculate the work done using the formula Work = Force × distance (stroke depth).

Performing these calculations:

Radius(r) = diameter / 2 = 2.50 cm / 2 = 1.25 cm = 0.0125 m

Area(A) = πr² = π × (0.0125 m)² = 4.9087 × 10⁻´ m²

Average force(F) = Pressure(P) × Area(A) = 2.40 × 10⁵ N/m² × 4.9087 × 10⁻´ m² = 117.81 N

Work(W) = Force(F) × distance(d) = 117.81 N × 0.30 m = 35.34 J

Thus, in one stroke, you do 35.34 Joules of work, and the average force you exert on the piston is 117.81 Newtons.

User Supun Sandaruwan
by
8.0k points