Answer:
(b) In ideal condition we neglect mass of spring but in real springs mass of spring adds another factor to its time period.
since we are adding a factor of mass to the system, and frequency being inversely proportional to squared root of mass, we can come to a general conclusion that it effectively reduces the natural frequency .
Explanation:
kindly check the attachment for explanation.
Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
![v^2=u^2+2as](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2as)
![v=\sqrt{523.69}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B523.69%7D)
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
Answer:
Force and displacement.
Explanation:
Work done is positive when we push table and it move in the direction of applied force.
1) The total mechanical energy of the rock is:
![E=U+K](https://tex.z-dn.net/?f=E%3DU%2BK)
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
![E_i=U=mgh](https://tex.z-dn.net/?f=E_i%3DU%3Dmgh)
where
![m=4 kg](https://tex.z-dn.net/?f=m%3D4%20kg)
is the mass,
![g=9.81 m/s^2](https://tex.z-dn.net/?f=g%3D9.81%20m%2Fs%5E2)
is the gravitational acceleration and
![h=5 m](https://tex.z-dn.net/?f=h%3D5%20m)
is the height.
Putting the numbers in, we find the potential energy
![U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J](https://tex.z-dn.net/?f=U%3Dmgh%3D%284%20kg%29%289.81%20m%2Fs%5E2%29%285%20m%29%3D196.2%20J)
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
![E_f=K= \frac{1}{2}mv^2](https://tex.z-dn.net/?f=E_f%3DK%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20)
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
![K=U=196.2 J](https://tex.z-dn.net/?f=K%3DU%3D196.2%20J)
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is: