Two blocks are connected by a light string that passes over a frictionless pulley as in the figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor.
(a) Assuming m1 > m2, find an expression for the speed of m1 just as it reaches the floor. (Use any variable or symbol stated above along with the following as necessary: g.)
(b) Taking m1 = 6.1 kg, m2 = 4.5 kg, and h = 3.1 m, evaluate your answer to part (a).
(c) Find the speed of each block when m1 has fallen a distance of 1.2 m.
(a) Since m1 > m2, m1 will fall freely while m2 will move upward. The only force acting on both blocks is gravity, so we'll use the following equation for motion under constant acceleration:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (0 in this case), a is the acceleration, and s is the distance traveled.
For m1, we can write:
v1^2 = 0^2 + 2a1h
where v1 is the final velocity of m1, a1 is the acceleration of m1, and h is the distance m1 falls.
For m2, we can write:
v2^2 = 0^2 + 2a2h
where v2 is the final velocity of m2 and a2 is the acceleration of m2.
Since the string connecting the blocks is inextensible, both blocks must have the same magnitude of acceleration, so we can write a1 = a2 = a.
The net force acting on m1 is m1g - T, where T is the tension in the string, and that on m2 is T - m2g. By applying Newton's second law (F = ma), we get:
m1g - T = m1a (1)
T - m2g = m2a (2)
Adding equations (1) and (2), we get:
m1g - m2g = m1a + m2a
a(m1 + m2) = g(m1 - m2)
a = g(m1 - m2) / (m1 + m2)
Now we can substitute this value of a in the equation for v1^2:
v1^2 = 2 * (g(m1 - m2) / (m1 + m2)) * h
v1 = sqrt[2 * (g(m1 - m2) / (m1 + m2)) * h]
This is the expression for the speed of m1 just as it reaches the floor.
(b) Now substituting the given values of m1, m2, and h:
m1 = 6.1 kg
m2 = 4.5 kg
h = 3.1 m
g = 9.81 m/s² (approximately)
v1 = sqrt[2 * (9.81(6.1 - 4.5) / (6.1 + 4.5)) * 3.1]
v1 = sqrt(2 * (9.81 * 1.6 / 10.6) * 3.1)
v1 ≈ 3.16 m/s
So, the speed of m1 just as it reaches the floor is approximately 3.16 m/s.
(c) To find the speed of each block when m1 has fallen a distance of 1.2 m, we can simply replace h with 1.2 m in the expression for v1:
v1 = sqrt[2 * (g(m1 - m2) / (m1 + m2)) * 1.2]
v1 = sqrt(2 * (9.81 * 1.6 / 10.6) * 1.2)
v1 ≈ 1.94 m/s
Since the blocks have the same magnitude of acceleration, their speeds will be equal at any given instant. Therefore, the speed of m2 will also be approximately 1.94 m/s when m1 has fallen a distance of 1.2 m.
To solve this problem, we will use the principles of conservation of energy.
(a) The system starts at rest, so the initial total mechanical energy is zero. As the blocks move, the potential energy of m1 decreases while the potential energy of m2 increases. This energy is transferred to kinetic energy.
The potential energy of m1 initially is given by m1 * g * h.
The potential energy of m2 initially is zero, as it is on the floor.
At the instant m1 reaches the floor, its potential energy becomes zero, and all the energy is converted to kinetic energy.
The kinetic energy of m1 can be calculated using the expression 1/2 * m1 * v^2, where v is the speed of m1 just as it reaches the floor.
Setting the initial potential energy of m1 equal to the final kinetic energy, we have:
m1 * g * h = 1/2 * m1 * v^2
Simplifying the equation, we get:
v^2 = 2 * g * h
Taking the square root of both sides, we find:
v = sqrt(2 * g * h)
(b) Plugging in the given values, m1 = 6.1 kg, m2 = 4.5 kg, h = 3.1 m, and using the acceleration due to gravity g = 9.8 m/s^2, we can evaluate the expression:
v = sqrt(2 * 9.8 m/s^2 * 3.1 m)
v ≈ 6.85 m/s
Therefore, the speed of m1 just as it reaches the floor is approximately 6.85 m/s.
(c) To find the speed of each block when m1 has fallen a distance of 1.2 m, we need to consider the conservation of mechanical energy for each block separately.
For m1, the initial potential energy is m1 * g * h, and the final potential energy is m1 * g * (h - 1.2). The difference in potential energy is converted to kinetic energy.
For m2, the initial potential energy is zero, and the final potential energy is m2 * g * 1.2. Since m2 is on the floor, its height does not change.
Setting the initial potential energy equal to the sum of the final potential energy and kinetic energy for each block, we have:
m1 * g * h = m1 * g * (h - 1.2) + 1/2 * m1 * v1^2
0 = m2 * g * 1.2 + 1/2 * m2 * v2^2
We have two unknowns here, v1 (velocity of m1) and v2 (velocity of m2). We need another equation. The key is that both blocks are connected by a light string passing through a pulley, which means the string does not slip or stretch. This implies that the string's length remains constant. Thus, the distance m1 falls is the same distance m2 rises.
Therefore, h - 1.2 = 1.2, or h = 2.4 m.
Now, we can solve the equations:
m1 * g * h = m1 * g * (h - 1.2) + 1/2 * m1 * v1^2
m2 * g * 1.2 + 1/2 * m2 * v2^2 = 0
Plugging in the given values, m1 = 6.1 kg, m2 = 4.5 kg, g = 9.8 m/s^2, and h = 2.4 m, we can solve these equations to find the velocities v1 and v2. However, the given values seem to be missing. Please provide the missing values, and I will be able to calculate the speeds.
To find the expression and evaluate the speed of m1, we need to understand the principles of mechanical energy conservation and motion equations.
(a) First, let's consider the initial and final states of the system. Initially, m1 is at a height h above the floor, and m2 is on the floor. Finally, when m1 reaches the floor, both blocks will be on the floor.
During the motion, the potential energy of m1 will be converted to the kinetic energy of m1. At the same time, the kinetic energy of m1 will be transferred to m2. Because there is no friction, the total mechanical energy of the system is conserved.
The potential energy of m1 at a height h is given by m1gh, where g is the acceleration due to gravity.
The kinetic energy of m1 just as it reaches the floor is given by (1/2)m1v^2, where v is the speed of m1.
Since the total mechanical energy is conserved, we can equate the initial and final energies:
m1gh = (1/2)m1v^2
Solving for v, we get:
v = sqrt(2gh/m1)
Therefore, the expression for the speed of m1 just as it reaches the floor is:
v = sqrt(2gh/m1)
(b) To evaluate the answer to part (a), we need to substitute the given values: m1 = 6.1 kg, m2 = 4.5 kg, h = 3.1 m, and g = 9.8 m/s^2.
v = sqrt(2 * 9.8 * 3.1 / 6.1) ≈ 4.36 m/s
The speed of m1 just as it reaches the floor is approximately 4.36 m/s.
(c) To find the speed of each block when m1 has fallen a distance of 1.2 m, we need to consider the conservation of mechanical energy separately for m1 and m2.
For m1:
The initial potential energy is m1gh.
The final potential energy is m1g(h - 1.2) since it has fallen by 1.2 m.
The final kinetic energy is (1/2)m1v1^2, where v1 is the speed of m1 after it has fallen 1.2 m.
Setting the initial and final energies equal, we have:
m1gh = m1g(h - 1.2) + (1/2)m1v1^2
Simplifying, we find:
v1^2 = 2g(1.2)
For m2:
Since it is on the floor, its final kinetic energy is (1/2)m2v2^2, where v2 is its speed.
Since there is no vertical motion for m2, there is no change in its potential energy.
Therefore, the equation is:
(1/2)m1v1^2 + (1/2)m2v2^2 = m1gh
Now, we need to solve these equations simultaneously to find the speeds v1 and v2.
Substituting the given values and solving the equations will provide the solution.