An inverted garbage can of Weight W is suspended in air by water from a geyser. The water shoots up from the ground with a speed u, at a constant rate dm/dt. Find the maximum height at which the garbage can rides.What assumption must be fulfilled for the maximum height to be reached?
Clue: If u=20m/s, W=10kg, dm/dt=0.5 kg/s, then hmax= 17m.
To find the maximum height at which the garbage can rides, we can use the principle of conservation of energy.
First, let's establish the assumptions for the problem:
1. The system is assumed to be ideal, with no losses due to friction, air resistance, or heat.
2. The gravitational force is assumed to be constant over the height range.
Now, let's break down the problem and find the maximum height.
Step 1: Calculate the initial potential energy and the final potential energy of the system.
The initial potential energy of the system is given by the weight of the garbage can (W) multiplied by the initial height (0), which is zero since the garbage can is at the ground level.
Initial potential energy = W * 0 = 0
The final potential energy of the system is given by the weight of the garbage can (W) multiplied by the final height (hmax).
Final potential energy = W * hmax
Step 2: Calculate the kinetic energy of the system.
The kinetic energy of the system is given by the kinetic energy of the water shooting up from the ground. Since the water is coming from a geyser with a constant rate of dm/dt, we can assume a constant velocity (u).
Kinetic energy = (1/2) * m * u^2
Where m is the mass of the water accumulated at height hmax. To find m, we need to consider the rate of change of mass (dm/dt) and the time interval t for which the water accumulates:
m = dm/dt * t
However, we don't have the value of t. We need to find it.
Step 3: Find the time (t) required for water accumulation.
We know that the velocity of water shooting up (u) and the rate of change of mass (dm/dt) are related by the equation:
u = (dm/dt) / A
Where A is the cross-sectional area of the jet. Since we don't have the value of A, we cannot directly calculate t.
However, we are given some clue values: u = 20 m/s, W = 10 kg, and dm/dt = 0.5 kg/s, with hmax = 17 m.
Let's assume that the clue values relate to the maximum height.
When the garbage can reaches its maximum height, all the water that was coming out during that time (dm/dt) is accumulated in the garbage can. Therefore, we can find t by dividing the mass of the water accumulated in the garbage can (dm/dt * t) by the mass flow rate (dm/dt).
t = (dm/dt * t) / dm/dt
Here, dm/dt cancels out, and we're left with t = t, which means t can be any value.
This implies that the minimum t required for the maximum height to be reached is 0, indicating an instantaneous accumulation of water.
Step 4: Calculate the final potential energy using the clue values.
Using the clue values, W = 10 kg, hmax = 17 m, and t = 0, we can find the final potential energy:
Final potential energy = W * hmax = 10 kg * 17 m = 170 J
Step 5: Equate the initial and final energies of the system.
Since energy is conserved, we can equate the initial and final potential energies:
0 = 170 J + [Kinetic energy]
Now, we need to find the kinetic energy.
Step 6: Calculate the kinetic energy using the clue value.
From the provided clue, hmax = 17 m, and the garbage can reaches the maximum height. Therefore, there is no kinetic energy at the maximum height, so [Kinetic energy] = 0 J.
0 = 170 J + 0 J
This equation is not balanced. Thus, the assumption given in the question that hmax = 17 m is not valid, and the maximum height cannot be reached.
To find the maximum height at which the garbage can rides, we can use the principle of conservation of energy.
The gravitational potential energy of the garbage can at its maximum height is equal to the work done by the water geyser.
Let's first find the work done by the water geyser:
The mass of water that is lifted per unit time is given by dm/dt.
So, in time dt, the mass of water lifted is dm = (dm/dt) * dt.
The work done by the water geyser in lifting this water is equal to the change in potential energy of the garbage can:
dW = m * g * dh,
Where m is the mass of the water lifted and g is the acceleration due to gravity.
Assuming that the garbage can is initially at rest, the kinetic energy at the maximum height is zero.
At its maximum height, the potential energy of the garbage can is given by:
ΔPE = m * g * h,
Where h is the maximum height reached by the garbage can.
Since the kinetic energy at the maximum height is zero, the total mechanical energy is conserved:
Initial potential energy + Initial kinetic energy = Final potential energy + Final kinetic energy
0 + 0 = m * g * h + 0
Simplifying, we get:
m * g * h = dW = m * g * dh
Canceling out the mass and acceleration due to gravity, we have:
h = dh
Integrating both sides of the equation, we get:
∫dh = ∫dh
Integrating with appropriate limits, we have:
h = hmax - h0,
Where h0 is the initial height of the garbage can before the water is released and hmax is the maximum height reached.
To find hmax, we need to know the values of u (water speed), W (garbage can weight), and dm/dt (rate of mass flow of water) and assume that the maximum height is reached.
Given that u = 20 m/s, W = 10 kg, dm/dt = 0.5 kg/s, and hmax = 17 m, we can substitute these values into the equation to find the value of h0:
hmax = h0 + ∫{u - (dm/dt)/A} dt,
Where A is the cross-sectional area of the opening at the bottom of the garbage can.
Simplifying,
17 = h0 + ∫{20 - (0.5)/A} dt
If we assume that the opening at the bottom of the garbage can is small enough so that the water escapes at a negligible velocity and the height of the garbage can doesn't change significantly during the time it takes for the water to reach the maximum height, then we can approximate the term (dm/dt)/A to zero, and the equation becomes:
17 = h0 + ∫20 dt
17 = h0 + 20t
Solving for h0, we get:
h0 = 17 - 20t
Since the garbage can is initially at rest, t = 0. Therefore, h0 = 17 - 0 = 17 m.
Therefore, the assumption that must be fulfilled for the maximum height to be reached is that the opening at the bottom of the garbage can is small enough so that the water escapes at a negligible velocity and the height of the garbage can doesn't change significantly during the time it takes for the water to reach the maximum height.
Your clue is impossible, water release at a velocity of 20m/s cannot lift a 10 kg garbage can under earth's gravity.
F=lambda 2v^2 (assuming elastic collision)
= 2mv^2/d
= 2(dm/dt)(v^2)/u
= 20N
which can lift a can of 20/9.81 kg
so your water can only lift a max of 2.04 kg garbage can.