A mass m1 = 3.3 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.4 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.87 m.
1) How much work is done by gravity on the two block system?
2) How much work is done by the normal force on m1?
3) What is the final speed of the two blocks?
4) How much work is done by tension on m1?
5) What is the tension in the string as the block falls?
1. work=m2*g*h
2. zero, if the table is horizontal
3. 1/2 (m1+m2)v^2=workin1.
4. work=1/2 m1 v^2
5. Tension*distance=workin4.
v^2=2W/(m1+m2)
How do you find the v^2?
1) Well, gravity has been working hard for years, so it's probably tired. But in this case, the work done by gravity on the two block system can be calculated using the formula: work = force * distance. Since the only force acting on the hanging mass is gravity, the work done on it is m2 * g * d, where m2 is the mass of the hanging block, g is the acceleration due to gravity, and d is the distance it falls.
2) The normal force on m1 does zero work because it acts perpendicular to the direction of displacement. It just stands there being normal, not doing any work.
3) To find the final speed of the two blocks, we'll need to consider conservation of energy. The potential energy lost by the hanging mass is equal to the kinetic energy gained by the system of blocks. Using the formula: m2 * g * d = (m1 + m2) * v^2 / 2, where v is the final speed of the two blocks, we can solve for v.
4) The tension in the string does negative work on m1 as it slows it down. The work done by tension can be calculated using the formula: work = force * distance. In this case, the distance is zero because there is no displacement in the direction of the tension force, so the work done by tension is zero.
5) The tension in the string can be found by considering the forces acting on the system. Since the system is in equilibrium, the tension in the string is equal to the weight of the hanging mass, which is m2 * g. So, just multiply the mass of the hanging block by the acceleration due to gravity to find the tension in the string.
To answer these questions, we can use several principles of physics, including the work-energy theorem and Newton's laws.
1) To find the work done by gravity on the two-block system, we need to calculate the change in gravitational potential energy. The formula for gravitational potential energy is given by U = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
In this case, the hanging mass m2 falls a distance d = 0.87 m. So, the change in potential energy for m2 is given by ΔU = m2 * g * d.
Since the potential energy of m1 doesn't change (as it is resting on a frictionless table), the work done by gravity on m1 is zero.
Therefore, the total work done by gravity on the two-block system is ΔU = (m1 + m2) * g * d. Plugging in the values, we get:
Work by gravity = (3.3 kg + 4.4 kg) * 9.8 m/s^2 * 0.87 m = 68.19 J
2) The normal force acts perpendicular to the surface of m1 and does no work since there is no displacement in the direction of the normal force. Therefore, the work done by the normal force on m1 is zero.
3) To find the final speed of the two blocks, we can use the principle of conservation of mechanical energy. The initial mechanical energy is equal to the final mechanical energy. We can calculate the initial mechanical energy as the sum of the potential energy and kinetic energy of the system.
Initially, the potential energy of m2 is m2 * g * h, where h is the initial height (which we assume to be zero since it hangs freely).
The initial kinetic energy of the system is zero because both blocks are initially at rest.
The final mechanical energy is the sum of the final potential energy and kinetic energy of the system. The final potential energy is m2 * g * d (since it falls a distance d). The final kinetic energy can be calculated as (1/2) * (m1 + m2) * v_f^2, where v_f is the final speed.
Equating the initial and final mechanical energies, we have:
0 + 0 = m2 * g * d + (1/2) * (m1 + m2) * v_f^2
Simplifying further, we get:
v_f^2 = 2 * g * d * (m2 / (m1 + m2))
Plugging in the values, we get:
v_f = √(2 * 9.8 m/s^2 * 0.87 m * (4.4 kg / (3.3 kg + 4.4 kg)))
4) The work done by tension on m1 can be calculated by finding the change in kinetic energy of m1. Since m1 starts at rest, the change in kinetic energy is given by the final kinetic energy of m1.
The final kinetic energy of m1 can be calculated as (1/2) * m1 * v_f^2.
Plugging in the values, we get:
Work by tension on m1 = (1/2) * 3.3 kg * v_f^2
5) The tension in the string can be calculated by considering the forces acting on m1. Since it is in equilibrium (not accelerating), the net force acting on it is zero.
The net force on m1 is the difference between the gravitational force pulling it down and the tension pulling it up. Therefore, we have:
Tension - m1 * g = 0
Solving for tension, we get:
Tension = m1 * g
Plugging in the values, we get:
Tension = 3.3 kg * 9.8 m/s^2