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 determine the speed of the masses (v) and the magnitude of tension in the cord (T), we can use the principle of conservation of energy. The total mechanical energy of the system remains constant throughout the motion.
Let's break down the problem step-by-step:
Step 1: Calculate the gravitational potential energy at the initial position for each mass.
Gravitational potential energy is given by U = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
For m1: U1 = m1gh
For m2: U2 = m2gh
Step 2: Calculate the final kinetic energy for each mass.
Kinetic energy is given by K = (1/2)mv^2, where m is the mass and v is the velocity.
For m1: K1 = (1/2)m1v^2
For m2: K2 = (1/2)m2v^2
Step 3: Apply the conservation of energy principle.
Since the only forms of energy involved are gravitational potential energy and kinetic energy, we can equate the initial potential energy to the final kinetic energy.
U1 + U2 = K1 + K2
Step 4: Substitute the known values and solve for v.
Given:
m1 = 4.40 kg
m2 = 80% of m1 (mass ratio m2/m1 = 0.8)
h = 1.00 m
Using substitution, we have:
m1gh + m2gh = (1/2)m1v^2 + (1/2)m2v^2
Substitute the mass ratio:
4.40 kg * 9.8 m/s^2 * 1.00 m + (0.8 * 4.40 kg) * 9.8 m/s^2 * 1.00 m = (1/2) * 4.40 kg * v^2 + (1/2) * (0.8 * 4.40 kg) * v^2
Simplifying the equation gives:
43.176 + 34.5408 = 2.20v^2 + 1.76v^2
Combine like terms:
77.7168 = 3.96v^2
Divide both sides by 3.96:
v^2 = 19.62
Take the square root of both sides:
v = √19.62
Using a calculator, we find v ≈ 4.43 m/s.
(a) The speed of the masses is approximately 4.43 m/s.
Step 5: Calculate the tension in the cord.
To find the tension (T), we can use either m1 or m2, as the tension in the cord is the same for both masses.
T = m1g - m1a
Given:
m1 = 4.40 kg
g = 9.8 m/s^2 (acceleration due to gravity)
a = acceleration of the masses
Since the masses start from rest and travel the same distance, their acceleration is the same. We can calculate the acceleration using the obtained value of v and the equation:
v^2 = u^2 + 2as
Where u is the initial velocity, s is the distance traveled, and a is the acceleration.
In this case, u = 0, s = 1.00 m, and v = 4.43 m/s.
(4.43 m/s)^2 = 0 + 2a(1.00 m)
Simplifying the equation gives:
19.62 m^2/s^2 = 2a
Divide both sides by 2:
a = 9.81 m/s^2
Now, we can substitute the obtained value of a into the equation for tension:
T = m1g - m1a
T = 4.40 kg * 9.8 m/s^2 - 4.40 kg * 9.81 m/s^2
Simplifying the equation gives:
T = 43.12 - 43.13
Calculating the result gives:
T ≈ -0.01 N
(b) The magnitude of the tension in the cord is approximately 0.01 N.