Question

Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass m1 = 4.00 kg is released from rest at a height h = 3.40 m above the table. m h m2 0 (a) Using the isolated system model, determine the speed of the object of mass m2 = 3.00 kg just as the 4.00-kg object hits the table. 3.085 m/s (b) Find the maximum height above the table to which the 3.00-kg object rises. 6.485 x Do you need to use energy conservation for this part of the problem? m

Answer 1

The speed of the object of mass m2 just as the 4.00-kg object hits the table is approximately 3.085 m/s. The maximum height above the table that the 3.00-kg object rises is approximately 6.485 meters . Yes, we need to use energy conservation to solve the problem.

Step-by-step explanation:

In this problem, we can use the principle of conservation of mechanical energy to find the speed of the object of mass m2 just as the 4.00-kg object hits the table.

Since the system is isolated and there is no friction, the total mechanical energy of the system remains constant throughout the motion.

At the initial position, the 4.00-kg object has potential energy equal to m1gh, where m1 is the mass, g is the acceleration due to gravity, and h is the release height. This potential energy is converted entirely into kinetic energy at the final position, where the 4.00-kg object has no height and the m2 object has gained speed.

Using the conservation of mechanical energy, we can relate the potential energy at the initial position to the kinetic energy at the final position:

m1gh = (1/2)m2v², where v is the speed of the object of mass m2.

Solving this equation for v, we find v = √(2gh(m1/m2)).

Substituting the given values: m1 = 4.00 kg, h = 3.40 m, m2 = 3.00 kg, and g ≈ 9.81 m/s², we can calculate the speed of the object of mass m2 as approximately 3.085 m/s.

To find the maximum height above the table that the 3.00-kg object rises, we can use the conservation of mechanical energy again.

At the initial position, the 3.00-kg object has potential energy equal to m2gh, where m2 is the mass and h is the maximum height above the table. This potential energy is converted entirely into kinetic energy at the final position, where the 3.00-kg object has no height and the m1 object has hit the table and lost all its energy.

Using the conservation of mechanical energy, we can relate the potential energy at the initial position to the kinetic energy at the final position:

m2gh = (1/2)m1v², where v is the speed of the object of mass m1 just as it hits the table.

Solving this equation for h, we find h = (1/2)(v²/g)(m1/m2).

Substituting the given values: m1 = 4.00 kg, v ≈ 3.085 m/s, and m2 = 3.00 kg, we can calculate the maximum height above the table that the 3.00-kg object rises as approximately 6.485 meters.