Answer:
Assuming air resistance is negligible, all of the potential energy that the object has at the top of the ramp is converted into kinetic energy by the time it gets to the bottom of the ramp. This is because no matter what path the object takes to move the 5m vertically (ie. falling straight down v. sliding on the ramp), gravity does the same amount of work on it.
Thus, calculate the total amount of potential energy at the top of the ramp:
Ep=mgh
Ep=4(9.81)5
Ep=196.2 Joules
Because all of this potential energy is converted into kinetic energy in the object by the bottom of the ramp, the object hits the spring with 196.2J of energy.
By using the formula for elastic potential energy, you can calculate exactly how far the spring compresses.
196.2=(1/2)k(x^2)
392.4=(350)(x^2)
1.1211=x^2
sqrt(1.1211)=x
x=1.059m
As for the last part of the question, after the object compresses the spring fully and stops momentarily, the spring converts it's elastic potential energy back into kinetic energy in the object and pushes it away again.
Explanation:
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Hi there!
Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.
We also know there was work done on the ball by air resistance that decreased the ball's total energy.
Let's do a summation using the equations:
Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)
Our final energy, since we set the zero-line to be at H2, is just kinetic energy.
And:
The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.
Therefore:
Solving for the work done by air resistance:
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that the friction force on two boxes is given as
Now we know by Newton's II law
so we have
Part b)
For block B we know that net force on it will push it forward with same acceleration so we have
Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as
