In the movement of the weight in vertical circle, using momentum balance, the largest tension is at the bottom of the circle. This is represented by:
<span>F = T - m g </span>
<span>T = F + m g
</span>F (centripetal) = mv^2/r
<span>= m v^2 / r + m g </span>
<span>m v^2 / r = T - m g </span>
<span>T= 0.5m * 100kgm/s^2 / 0.2kg - 9.81m/s^2 * 0.5m </span>
<span>T= 245 m^2/s^2 </span>
Answer:
<u></u>
- <u>1. The potential energy of the swing is the greatest at the position B.</u>
- <u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>
Explanation:
Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.
The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>
Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
Answer:
a = 6.53 m/s^2
v = 11.5689 m/s
Explanation:
Given data:
engine power is 217 hp
70 % power reached to wheel
total mass ( car + driver) is 1530 kg
from the data given
2/3 rd of weight is over the wheel
w = 2/3rd mg
maximum force
we know that F = ma
the new power is 
solving for speed v
![v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}](https://tex.z-dn.net/?f=v%20%3D%200.7%20%5Cfrac%7B217%20%5B%5Cfrac%7B746%20w%7D%7B1%20hp%7D%5D%7D%7B1500%20%5Ctimes%206.53%7D)
v = 11.5689 m/s
S=56, u=0, v=33, a=?, t=3.4
v=u+at
33=3.4 a
a = 9.7m/s^2
