In the frictionless apparatus m1=1.7kg a)What is m2 if both masses are at rest? b) How about if both masses are moving at constant velocity?
Without a figure or a description of the apparatus, help cannot be provided.
To determine the value of m2 in both scenarios, let's consider the forces acting on the system and apply Newton's laws of motion.
a) When both masses are at rest:
In a frictionless apparatus, the only force acting on the system is the force due to the tension in the string connecting the masses. Since the masses are at rest, we know that the net force on the system is zero.
Let T be the tension in the string. For mass m1, we have:
m1 * g - T = 0, where g is the acceleration due to gravity.
m1 * g = T
Similarly, for mass m2 which is at rest, we also have:
m2 * g - T = 0
m2 * g = T
Since both sides of the equation are equal to T, we can equate the two expressions:
m1 * g = m2 * g
Now, we can cancel out the acceleration due to gravity on both sides of the equation:
m1 = m2
Therefore, when both masses are at rest, m2 will have the same value as m1, which is 1.7 kg.
b) When both masses are moving at constant velocity:
In this scenario, the system is in equilibrium. This means that the net force acting on the system is still zero.
Again, the only force acting on the system is the tension in the string. Therefore, we have:
m1 * g - T = 0
m1 * g = T
For mass m2, we have:
m2 * g - T = 0
m2 * g = T
Since we are now considering the masses to be moving at constant velocity, we can conclude that T is equal to the frictional force acting on the system. Let's call it Ffriction.
Therefore, we can write the equation:
m1 * g = Ffriction
m2 * g = Ffriction
Since we know that both forces are equal, we can equate the two expressions:
m1 * g = m2 * g
Canceling out the acceleration due to gravity on both sides of the equation, we get:
m1 = m2
Hence, when both masses are moving at constant velocity, m2 will have the same value as m1, which is 1.7 kg.