Ball rolls down ramp, ball is released from spring, ball drops down chute.
Answer:
a. 652.68N
b. -2349.65J
c. -3116.12J
d. 5465.77J
e. Zero
Explanation:
a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).
Fg = Fm + Fr
Fm = Fg - Fr
Fm = mgsin(28°) - umgcos(28°)
u = coefficient of frictional force.
Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28
Fm = 1518.27 - 865.59
Fm = 652.68N
b. Work done by man is:
Wm = -Fm * d
Wm = -652.68 * 3.6
Wm = -2349.65J
c. Work done by friction force:
W(Fr) = -Fr * d
W(Fr) = -865.59 * 3.6
W(Fr) = -3116.12J
d. Work done by gravity:
Wg = Fg * d
Wg = 1518.27 * 3. 6
Wg = 5465.77J
e. Net work done on the piano is:
Work done by friction + work done by gravity + work done by man
= -3116.12 + 5464.77 + (-2349.65)
= 0J
Answer:
In 2.748 sec the mailbag reached the ground
Explanation:
We have given height from the ground ![h=3.25t^3](https://tex.z-dn.net/?f=h%3D3.25t%5E3)
At t =2.25 sec helicopter releases a small mailbag so at t = 2.25 sec height from the ground ![h=3.25t^3=3.25\times 2.25^3=37.01m](https://tex.z-dn.net/?f=h%3D3.25t%5E3%3D3.25%5Ctimes%202.25%5E3%3D37.01m)
When the mail box is drooped its initial velocity would zero so u = 0 m/sec
Acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
According to third law of motion ![h=ut+\frac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![37.01=0\times t+\frac{1}{2}\times 9.8\times t^2](https://tex.z-dn.net/?f=37.01%3D0%5Ctimes%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%20t%5E2)
![t^2=7.553](https://tex.z-dn.net/?f=t%5E2%3D7.553)
t = 2.748 sec
When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy
![U = mgh](https://tex.z-dn.net/?f=U%20%3D%20mgh)
so when height of the car becomes half then its potential energy is given as
![U_f = \frac{mgh}{2}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7Bmgh%7D%7B2%7D)
so final potential energy when car falls down by half of the height will become half of the initial potential energy
So it is U = 50 MJ after falling down
Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car
So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ
so we can say at this point magnitude of potential energy and kinetic energy will be same
<em>A. the same as the potential energy at that point.</em>
For those seeking for the answer, its a source of electrical energy.