The two masses shown in the figure are each initially 1.80 m above the ground, and the massless frictionless pulley is fixed 4.8 m above the ground. (mA = 1.5 kg and mB = 3.4 kg.) What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its "launch" speed. Assume the mass doesn't hit the pulley. Ignore the mass of the cord.]
To find the maximum height that the lighter object reaches after the system is released, we need to consider the conservation of energy.
First, let's determine the acceleration of the lighter mass (mA). We can use Newton's second law and the tension in the rope to find the acceleration:
Tension = mA * acceleration
The tension in the rope is also equal to the weight of the heavier mass (mB) since the system is in equilibrium:
Tension = mB * g
where g is the acceleration due to gravity.
Substituting the values into the equations, we get:
mB * g = mA * acceleration
Rearranging the equation, we have:
acceleration = (mB * g) / mA
Next, we'll find the velocity of the lighter mass (mA) at the moment the heavier mass (mB) hits the ground. We can use the equation of motion:
v^2 = u^2 + 2*a*s
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since both masses start from rest, the initial velocity (u) is 0, and the distance traveled (s) is the height of the pulley (4.8 m).
Plugging in the values, we get:
v^2 = 0 + 2 * acceleration * 4.8
The velocity (v) at this point will be the "launch" speed of the lighter mass.
Now, using the conservation of energy, we can find the maximum height reached by the lighter object.
The initial gravitational potential energy of the lighter mass is given by:
PE_initial = mA * g * initial height
The final gravitational potential energy at the maximum height reached is given by:
PE_final = mA * g * final height
Considering that the initial kinetic energy of the lighter mass is converted into gravitational potential energy at the maximum height, we have:
KE_initial = PE_final
Solving for the final height, we get:
final height = (KE_initial) / (mA * g)
The initial kinetic energy (KE_initial) can be calculated using the kinetic energy formula:
KE_initial = (1/2) * mA * v^2
Plugging in the values and calculating step-by-step, we can find the maximum height reached by the lighter object.
To find the maximum height reached by the lighter object, we need to determine its launch speed.
First, let's find the acceleration of the lighter mass (mA = 1.5 kg). We can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the difference between the gravitational force and the tension in the string.
The gravitational force on mA is given by:
F_gravity = m * g
where m is the mass and g is the acceleration due to gravity. Given that g is approximately 9.8 m/s^2, the gravitational force on mA is:
F_gravity = 1.5 kg * 9.8 m/s^2 = 14.7 N
The tension in the string can be found by considering the heavier mass (mB = 3.4 kg). The net force on mB is the sum of the tension in the string and the gravitational force acting on mB:
F_net = T - mB * g
Since the system is released and the heavier mass is allowed to fall, it will accelerate downward. At the same time, the lighter mass will accelerate upward. At the moment the heavier mass hits the ground, the lighter mass will reach its launch speed. At that point, the net force acting on it will be zero.
So, we have:
F_net = 0
T - mB * g = 0
Substituting the values we have:
T - 3.4 kg * 9.8 m/s^2 = 0
T = 33.32 N
Now, we can calculate the acceleration of the lighter mass:
F_net = T - F_gravity
0 = T - m * g
0 = 33.32 N - 1.5 kg * 9.8 m/s^2
0 = 33.32 N - 14.7 N
0 = 18.62 N
The acceleration is given by:
a = F_net / m
a = 18.62 N / 1.5 kg
a ≈ 12.41 m/s^2
Now that we have the acceleration, we can calculate the launch speed of the lighter mass. At the moment the heavier mass hits the ground (after traveling a distance of 4.8 m), the lighter mass will reach its maximum height. The final velocity (vf) of the lighter mass at that point can be found using the equation:
vf^2 = vo^2 + 2aΔh
where vo is the initial velocity (which is zero because it starts from rest), a is the acceleration, and Δh is the displacement.
We know Δh = 4.8 m, so the equation becomes:
vf^2 = 0^2 + 2 * 12.41 m/s^2 * 4.8 m
Simplifying:
vf^2 = 118.78 m^2/s^2
Taking the square root of both sides:
vf = sqrt(118.78 m^2/s^2)
vf ≈ 10.91 m/s
Therefore, the launch speed of the lighter mass is approximately 10.91 m/s.
To find the maximum height reached by the lighter mass, we can use the kinematic equation:
vf^2 = vo^2 + 2aΔh
Since the lighter mass is at its maximum height, its final velocity (vf) is zero. The initial velocity (vo) is 10.91 m/s, and the acceleration (a) is -9.8 m/s^2 (due to the force of gravity acting downward). Substituting these values into the equation:
0^2 = 10.91^2 + 2 * -9.8 m/s^2 * Δh
Simplifying:
0 = 119.32 - 19.6 m/s^2 * Δh
Rearranging the equation to solve for Δh:
19.6 m/s^2 * Δh = 119.32
Δh = 119.32 / 19.6 m/s^2
Δh ≈ 6.08 m
Therefore, the maximum height reached by the lighter object is approximately 6.08 m above the ground.