A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.8 m/s and the tension in the rope is T = 21.7 N.

1)Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg.
How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)?
2)Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

1)

(1/2)mv^2 = mgh, where h is 4/5 of the height

2)
T=ma, where a = v^2/r, v being answer from part 1 and r is 1/5 of length of rope

26.73 m/s

To find the answer to these questions, we can use the principles of conservation of energy and the centripetal force acting on the mass in the circular motion.

1) To determine the velocity of the mass at the same height as the peg, we can use the conservation of mechanical energy. At the bottom of the pendulum's path, all the potential energy is converted into kinetic energy. At its vertical height, the mass only has potential energy. Therefore, we can equate the initial kinetic energy to the potential energy at the vertical position.

Using the equation for kinetic energy (K.E. = 0.5 * mass * velocity^2) and potential energy (P.E. = mass * g * height) where g is the acceleration due to gravity, we can write:

0.5 * mass * velocity^2 = mass * g * height

Since height is the distance from the bottom to the vertical position, we can call it h.

Solving for velocity, we get:

velocity = sqrt(2 * g * h)

Substituting the given values of v = 2.8 m/s and g = 9.8 m/s^2, we can calculate the height (h) using the formula above and then plug it back into the equation to find the velocity at the same height as the peg.

2) To find the tension in the rope at the same vertical height as the peg, we can consider the forces acting on the mass at that point. The tension in the rope and the gravitational force are the only forces acting on the mass in the vertical direction.

Since the velocity of the mass is not changing, we know that the net force in the vertical direction is zero. When the rope is at the same height as the peg, the gravitational force is pulling the mass downward, and the tension in the rope is pulling it upward. Therefore, we can equate these two forces:

Tension = gravitational force

Using the equation for gravitational force (Fg = mass * g), we can write:

Tension = mass * g

Plugging in the given value of T = 21.7 N and the mass of the object, we can calculate the tension in the rope at the same vertical height as the peg.

A person throws a stone straight upward at an initial speed of 15 m/s on a bridge that is 25m above the surface of the water. If the stone just misses the bridge on the way down, What is the speed of the stone just before it hits the water?

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 32.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. What is the speed of the rock just before it hits the street? How much time elapses from when the rock is thrown until it hits the street?