A pendulum of mass .9 kg and length 1.86 meters is initially displaced .3906 cm in the horizontal direction from its equilibrium position. The pendulum is released from rest at this position.
How much work does gravity therefore do on the pendulum during its fall to the equilibrium position, and how much work does the pendulum do against gravity?
OK can you use this test question to explain to me the difference in between work done BY gravity and work done against gravity?
Note that to solve this problem you need to use right triangles to figure out how far the pendulum descends from release to the equilibrium position.
To answer your question about work done by vs. on the system:
The pendulum moves lower as it returns to its equilibrium position. So its displacement is in the same direction as the gravitational force exerted ON the pendulum, and the work done by that force is therefore positive. So gravity does positive work ON the pendulum.
At the same time the pendulum exerts an upward gravitational force on the Earth. That force is in the direction opposite the displacement of the descending pendulum and the work done by that force is therefore negative. This is the force exerted by the pendulum and is the work done BY the pendulum.
Positive work done ON the pendulum or negative work done BY the pendulum tend to increase its KE.
A mass of .62 kg rests on a frictionless tabletop, attached by a string running horizontally to and then over a pulley to a mass of .1488 kg.
When the system is released what will be its acceleration?
What is the tension in the strings?
Can you work this one for me. I have no idea how to get the tension of the string?
The normal force balances the gravitational force acting on the .62 kg mass. The net force on the system is thereforem just the 9.8 m/s^2 * .1488 kg = 1.46 N exerted by gravity on the hanging mass.
Both the .62 kg and the .1488 kg masses are accelerated by this net force, so the acceleration is
a = Fnet / m = 1.46 N / (.62 kg + .1488 kg) = 1.46 N / (.76 kg) = 1.96 m/s^2 approx..
It is the tension in the string that provides the force that accelerates the .62 kg mass. So the tension must be .62 kg * 1.96 m/s^2, approx..
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 13 cm/s^2. .179 seconds later it passes a point 5 cm up the ramp from its initial position.
What are its possible initial velocities, and what is a possible scenario for each?
What is v0 for this .179 sec interval?
What is a (caution: you are given the magnitude of a and you have to figure out whether a is positive or negative)?
What is `ds?
What therefore are v0 and vf?
What is the maximum distance the object travels up the ramp?
This is a new problem that, of the three quantities given in the original situation, uses only the acceleration.
It also uses the initial velocity you found above.
Max distance is not reached until the object comes to rest for an instant before beginning to accelerate back down the ramp.
So the acceleration situation between start and max distance has v0 equal to the initial velocity you just found, vf equal to 0 and the same acceleration as before.
What therefore at `dt and `ds?
I've seen a few questions like this one but I couldn't ever figure it out, can you help me?