Two blocks are connected by massless string that is wrapped around a pulley. Block 1 has a mass m1=6.90 kg, block 2 has a mass m2=2.30 kg, while the pulley has a mass of 1.70 kg and a radius of 12.2 cm. The pulley is frictionless, and the surface mass 1 is on is also frictionless.
If the blocks are released from rest, how far will block 2 fall in 2.00 s?
To find the distance block 2 will fall in 2.00 s, we need to analyze the motion of the system.
First, let's determine the acceleration of the system. Since the pulley is frictionless, the tension in the string will be the same on both sides of the pulley.
Using Newton's second law, we can write down the equation for block 1:
m1 * a = T, where T is the tension in the string.
Similarly, for block 2:
m2 * a = T', where T' is also the tension in the string.
Since the blocks are tied together by the string, we can relate the tensions T and T' using the pulley. As the blocks move, the string unwinds from one side of the pulley and winds onto the other side. Since the pulley has a mass, we need to consider its inertia.
The acceleration of the pulley can be found using the equation:
mP * aP = T - T',
where mP is the mass of the pulley and aP is its acceleration.
However, we are interested in the acceleration of the blocks, so we need to express the acceleration of the pulley in terms of the acceleration of the blocks.
The linear acceleration a of the blocks is related to the angular acceleration α of the pulley by the equation:
a = α * R,
where R is the radius of the pulley.
We can also relate the angular acceleration of the pulley to its linear acceleration using the equation:
aP = -α * R,
since the pulley moves in the opposite direction compared to the blocks.
Combining the equations, we have:
mP * (-α * R) = T - T'
mP * R * a / R = T - T'
mP * a = T - T'
Now let's consider the tension T. Since block 1 is moving downward and block 2 is moving upward, the net force on block 1 is:
F1 = m1 * g - T = m1 * a,
where g is the acceleration due to gravity.
Similarly, the net force on block 2 is:
F2 = T' - m2 * g = m2 * a.
Simplifying these equations, we have:
T = m1 * (g - a),
T' = m2 * (g + a).
Now we can substitute these values back into the equation for mP * a = T - T':
mP * a = m1 * (g - a) - m2 * (g + a),
mP * a = (m1 - m2) * g - (m1 + m2) * a,
mP * a + (m1 + m2) * a = (m1 - m2) * g,
a * (mP + m1 + m2) = (m1 - m2) * g,
a = [(m1 - m2) * g] / (mP + m1 + m2).
Now we have the acceleration of the system. To find the distance block 2 will fall in 2.00 s, we can use the kinematic equation:
d = v0*t + (1/2)*a*t^2,
where v0 is the initial velocity which is zero, t is the time which is 2.00 s, and a is the acceleration.
Plugging in the values, we have:
d = (1/2)*a*t^2,
d = (1/2)*[(m1 - m2) * g] / (mP + m1 + m2) * (2.00 s)^2.
Now you can calculate the value of d using the given values for m1, m2, mP, g, and t.