A system has a mass of 600kg and a cross section area of 0.08m². Te piston on the right has a cross sectional area of 0.0025m² and a negligible weight. The fluid in the apparatus has oil with a density of 780kg/m³. (g=10m/s²)
We are given:
- Mass of system (m): 600kg
- Cross-sectional area of system (A): 0.08m²
- Cross-sectional area of piston on the right (a): 0.0025m²
- Density of oil (ρ): 780kg/m³
- Acceleration due to gravity (g): 10m/s²
To find the force exerted on the piston:
1. Determine the pressure of the oil:
The pressure exerted by a fluid at a given depth is given by:
P = ρgh
where P is pressure, ρ is density, g is acceleration due to gravity, and h is the depth.
Since the apparatus is vertical, we can assume that the depth is equal to the height of the system (h = 1m). Therefore, the pressure exerted by the oil on the system is:
P = ρgh = (780kg/m³)(10m/s²)(1m) = 7800 Pa
2. Determine the force exerted on the system:
The force exerted on a surface by a fluid is given by:
F = PA
where F is the force, P is pressure, and A is the area of the surface.
Therefore, the force exerted by the oil on the bottom of the system is:
F = PA = (7800 Pa)(0.08m²) = 624 N
3. Determine the force exerted on the piston:
The force exerted by the oil on the piston is given by:
f = pa
where f is the force, p is the pressure, and a is the area of the piston.
Therefore, the force exerted by the oil on the piston is:
f = pa = (7800 Pa)(0.0025m²) = 19.5 N
4. Determine the net force and acceleration of the system:
Since the force exerted on the system (624 N) is greater than the force exerted by the oil on the piston (19.5 N), there is a net force of:
Fnet = 624 N - 19.5 N = 604.5 N
The acceleration of the system is given by:
a = Fnet/m = 604.5 N/600kg = 1.0075 m/s²
Therefore, the system has an acceleration of 1.0075 m/s².