Two blocks are arranged as shown. The pulley can be considered to be massless, and friction is negligible. M1 is four times more massive than M2.
M1--->0(Pulley)
V
V (down direction)
V
M2
If the system is released from rest, how far will M1 travel in 0.523 s?
M2 is supposed to be hanging down from the pulley
The pulling force is M2*g
The moving mass is (M1+M2)
F=ma
M2*g=(M1+M2)a
solve for a
d=1/2 a t^2
To determine how far M1 travels in 0.523 s, we can use the equation for the displacement of an object under constant acceleration:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time
Since the system is released from rest, the initial velocity (u) of M1 is 0. The acceleration (a) of both blocks can be calculated using the concept of an inclined plane.
The force acting on M1 is the weight of M1 (W1 = m1 * g), where g is the acceleration due to gravity. The force acting on M2 is the weight of M2 (W2 = m2 * g).
The net force acting on the system is (W1 - W2). As per the problem statement, M1 is four times more massive than M2. Therefore, M2 experiences 4 times more force than M1.
Hence, the net force (Fnet) is 3/4 times the weight of M2.
To find the acceleration (a), we can use Newton's second law: Fnet = ma
Therefore, a = Fnet / m1
Now, let's calculate the acceleration (a):
Fnet = (3/4) * W2
= (3/4) * (m2 * g)
= (3/4) * (m2 * 9.8) [Assuming g = 9.8 m/s^2]
a = Fnet / m1
= ((3/4) * (m2 * 9.8)) / m1
The distance traveled by M1 in time t can be calculated using the equation:
s = ut + (1/2)at^2
Since the initial velocity (u) is 0, the equation simplifies to:
s = (1/2)at^2
Now, we can substitute the values:
s = (1/2) * (((3/4) * (m2 * 9.8)) / m1) * t^2
Let's assume values for m1 and m2, and calculate the distance traveled by M1 in 0.523 s.
To find out how far M1 will travel in 0.523 seconds, we can use the equations of motion. In this system, M1 is connected to the pulley, and M2 is hanging freely.
Let's assume that the acceleration of the system is "a" and the tension in the string connecting M1 and M2 is "T."
1. First, let's find the acceleration of the system. Since M1 is four times more massive than M2, we can write the equation for the net force:
M1 * a = M1 * g - T
Where M1 * g is the gravitational force acting on M1, and T is the tension in the string. Notice that there is a negative sign because the tension is acting in the opposite direction to the gravitational force.
2. Since the pulley is massless, the tension in the string is the same throughout the system. So we can also write the equation for M2:
M2 * a = M2 * g + T
Where M2 * g is the gravitational force acting on M2, and T is the tension in the string. Notice that there is a positive sign because the tension is acting in the same direction as the gravitational force.
3. Now, let's eliminate the tension in the equations by adding the two equations together:
M1 * a + M2 * a = (M1 + M2) * g
Since M1 is four times more massive than M2, we can substitute M1 = 4 * M2 into the equation:
(4 * M2) * a + M2 * a = (4 * M2 + M2) * g
Combining like terms:
5 * M2 * a = 5 * M2 * g
Dividing both sides by 5 * M2:
a = g
So the acceleration of the system is equal to the acceleration due to gravity.
4. Now that we know the acceleration, we can use the equation of motion to find the distance traveled by M1. The equation is:
S = u * t + 0.5 * a * t^2
Since the system is released from rest, the initial velocity (u) of M1 is 0. Therefore, the equation simplifies to:
S = 0.5 * a * t^2
Substituting the known values:
S = 0.5 * g * t^2
Plugging in the values of g = 9.8 m/s^2 and t = 0.523 s, we can calculate the distance traveled by M1.