A 83-kg worker clings to a lightweight rope going over a lightweight, low-friction pulley. The other end of the rope is connected to a 66-kg barrel of bricks. If the worker is initially at rest 16 m above the ground, how fast will he or she be moving when he or she hits the ground?
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*16 = 313.6,
Vf = 17.7m/s.
To determine the speed at which the worker will hit the ground, we can use the principle of conservation of energy. The total mechanical energy of the system (worker and barrel) remains constant throughout.
First, let's calculate the gravitational potential energy of the worker at a height of 16 m above the ground. The formula for gravitational potential energy is given by:
Potential Energy = mass * gravity * height
The mass of the worker is 83 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the height is 16 m. Plugging these values into the equation, we get:
Potential Energy = 83 kg * 9.8 m/s^2 * 16 m = 128,294 J
Now let's calculate the gravitational potential energy of the barrel. Using the same formula:
Potential Energy = mass * gravity * height
The mass of the barrel is 66 kg, the acceleration due to gravity is 9.8 m/s^2, and the height is also 16 m. Plugging these values into the equation, we get:
Potential Energy = 66 kg * 9.8 m/s^2 * 16 m = 102,528 J
Since the total mechanical energy remains constant, we can equate the potential energy of the worker to the kinetic energy of the system when they reach the ground.
Kinetic Energy = Total Mechanical Energy - Potential Energy
The kinetic energy of the system is given by the formula:
Kinetic Energy = 0.5 * total mass * velocity^2
The total mass of the system is the sum of the masses of the worker and the barrel, which is 83 kg + 66 kg = 149 kg.
Plugging the values into the kinetic energy equation, we get:
0.5 * 149 kg * velocity^2 = (128,294 J + 102,528 J)
Simplifying the equation, we have:
0.5 * 149 kg * velocity^2 = 230,822 J
Now we can solve for the velocity. Dividing both sides of the equation by 0.5 * 149 kg, we get:
velocity^2 = (230,822 J) / (0.5 * 149 kg)
velocity^2 = 3097.62 m^2/s^2
Taking the square root of both sides, we find:
velocity = sqrt(3097.62 m^2/s^2) = 55.68 m/s
Therefore, the worker will be moving at a speed of approximately 55.68 m/s when he or she hits the ground.