Two particles of masses 6kg and 3kg are attached to the two ends of an inextensible string passing over a pulley.the string is kept taut and the masses are released.. Calculate the common velocity of the particle and the distancetravelled after 5s
To solve this problem, we can use the principles of Kinematics and Newton's laws of motion.
First, let's analyze the forces acting on each mass:
Mass 1 (6 kg): The downward force due to gravity is given by F1 = m1 * g, where m1 is the mass of the first particle (6 kg) and g is the acceleration due to gravity (9.8 m/s^2). As the mass descends, the tension in the string pulls it upwards, balancing out the force due to gravity. So, the net force acting on mass 1 is T - F1 = m1 * a, where T is the tension in the string and a is the acceleration.
Mass 2 (3 kg): The downward force due to gravity is given by F2 = m2 * g, where m2 is the mass of the second particle (3 kg). Since this mass is moving upwards, the tension in the string is helping to lift it. So, the net force acting on mass 2 is T + F2 = m2 * a.
Since the string is inextensible, the magnitudes of the accelerations of the two masses must be equal, i.e., a = -a.
To find the magnitude of the common acceleration, we can solve the system of equations:
T - F1 = m1 * a
T + F2 = m2 * (-a)
Substituting the values, we have:
T - (6 kg * 9.8 m/s^2) = 6 kg * a
T + (3 kg * 9.8 m/s^2) = 3 kg * (-a)
Simplifying these equations we have:
T - 58.8 = 6a
T - 29.4 = -3a
Adding the two equations together:
2T - 88.2 = 0
Solving for T, we find:
T = 44.1 N
Now, we can find the magnitude of the common acceleration (a):
6a = T - 58.8
6a = 44.1 - 58.8
6a = -14.7
a = -2.45 m/s^2
Since the masses are released together, they have been accelerating for 5 seconds. We can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 since they were at rest), a is the acceleration, and t is the time (5 seconds).
Using this equation, we have:
v = 0 + (-2.45 m/s^2) * 5 s
v = -12.25 m/s
The negative sign indicates that the velocity is in the opposite direction of the initial motion.
To find the distance traveled, we can use the kinematic equation:
s = ut + (1/2)at^2
where s is the distance traveled.
Using this equation, we have:
s = 0 * 5 + (1/2) * (-2.45 m/s^2) * (5 s)^2
s = 0 + (1/2) * (-2.45 m/s^2) * 25 s^2
s = 0 + (-30.625 m)
s = -30.625 m
Again, the negative sign indicates that the displacement is in the opposite direction of the initial motion.
So, after 5 seconds, the common velocity of the particles is -12.25 m/s and the distance traveled (displacement) is -30.625 m.
Note: The negative sign indicates the direction of motion, where positive means upward and negative means downward.