As shown in the figure below, two masses m1 = 4.40 kg and m2 which has a mass 80.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If m1 and m2 start from rest, after they have each traveled a distance h = 1.00 m, use energy content to determine the following.
(a) speed v of the masses
m/s
(b) magnitude of the tension T in the cord
N
To solve this problem, we will use the principle of conservation of energy. The potential energy of the system at the initial position will be converted to kinetic energy when the masses reach their final position.
First, let's calculate the potential energy at the initial position (h = 0) for the system. The potential energy is given by the formula:
PE_initial = m1 * g * h
where m1 is the mass of the first object, g is the acceleration due to gravity, and h is the height.
Given that m1 = 4.40 kg and h = 1.00 m, and the acceleration due to gravity is approximate 9.8 m/s², we can calculate:
PE_initial = 4.40 kg * 9.8 m/s² * 1.00 m = 43.12 J
Since the masses start from rest and there is no friction, the total mechanical energy of the system remains constant. Therefore, the potential energy at the initial position (PE_initial) will be equal to the sum of the final kinetic energy for both masses (KE_final):
PE_initial = KE_final
Now, let's calculate the final kinetic energy (KE_final) for both masses. The kinetic energy is given by the formula:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
For m1, the final kinetic energy can be calculated as:
KE_m1 = (1/2) * m1 * v1^2
For m2, since its mass is 80.0% of m1, we can calculate its mass:
m2 = 0.80 * m1
And the final kinetic energy for m2 can be calculated as:
KE_m2 = (1/2) * m2 * v2^2
Since the masses are connected by a cord, their velocities are the same. Therefore, v1 = v2 = v.
Now, let's substitute the values we have:
PE_initial = KE_m1 + KE_m2
43.12 J = (1/2) * 4.40 kg * v^2 + (1/2) * 0.80 * 4.40 kg * v^2
43.12 J = (1/2) * 4.40 kg * v^2 + (1/2) * 0.80 * 4.40 kg * v^2
43.12 J = (1/2) * (4.40 kg + 0.80 * 4.40 kg) * v^2
43.12 J = (1/2) * 4.40 kg * (1 + 0.80) * v^2
43.12 J = (1/2) * 4.40 kg * (1.80) * v^2
43.12 J = 3.96 kg * v^2
Now, we can solve this equation for v, the speed of the masses:
v^2 = (43.12 J) / (3.96 kg)
v^2 = 10.899 J/kg
v ≈ √(10.899 J/kg)
v ≈ 3.30 m/s
So, the speed of the masses is approximately 3.30 m/s.
Next, let's calculate the tension in the cord. The tension (T) in the cord can be derived from the net force acting on the system. At the final position, the net force can be calculated using Newton's second law:
net force = m1 * a
where a is the acceleration of the system.
Since the masses are connected by a cord, their acceleration is the same. Therefore, a1 = a2 = a.
Now, let's calculate the net force. The net force is given by the formula:
net force = T - m1 * g
where T is the tension in the cord and g is the acceleration due to gravity.
Substituting the values we have:
m1 * a = T - m1 * g
4.40 kg * a = T - 4.40 kg * 9.8 m/s²
To find the magnitude of the tension T, we need to solve for T. Rearranging the equation:
T = 4.40 kg * a + 4.40 kg * 9.8 m/s²
Now, we need to find the value of a. The acceleration a can be calculated using the kinematic equation:
v^2 = u^2 + 2as
where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the displacement.
Given that the masses start from rest (u = 0), the final velocity v is 3.30 m/s (as calculated earlier), and the displacement s is 1.00 m, we can solve for a:
v^2 = u^2 + 2as
(3.30 m/s)^2 = 0 + 2a * 1.00 m
10.89 m^2/s^2 = 2a
Simplifying further:
2a = 10.89 m^2/s^2
a = (10.89 m^2/s^2) / 2
a ≈ 5.44 m/s²
Now, substituting the value of a into the equation for T:
T = 4.40 kg * 5.44 m/s² + 4.40 kg * 9.8 m/s²
T = 23.94 N + 43.12 N
T ≈ 67.06 N
Therefore, the magnitude of the tension in the cord is approximately 67.06 N.