If two charges are moved so that the distance between them is reduced
by half, then the forces between them become 4 times as strong.
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
![vf^2 = vi^2 - 2*g*x](https://tex.z-dn.net/?f=vf%5E2%20%3D%20vi%5E2%20-%202%2Ag%2Ax)
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
![0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s](https://tex.z-dn.net/?f=0%5E2%20%3D%20vi%5E2%20-%202%2A%28%209.80%20%29%2A%28%201.40%20%29%5C%5C%5C%5Cvi%5E2%20%3D%2027.44%5C%5C%5C%5Cvi%20%3D%20%5Csqrt%7B27.44%7D%20%3D%205.24%20m%2Fs)
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
![E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J](https://tex.z-dn.net/?f=E_k%20%3D%200.5%2Am%2A%28vi%29%5E2%5C%5C%5C%5CE_k%20%3D%200.5%2A0.86%2A27.44%5C%5C%5C%5CE_k%20%3D%2011.8%20J)
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
Part A.
To get the acceleration of the system we consider the two blocks as a single mass. For this situation we have, from Newton's second law, that:
![T-W=(m_1+m_2)a](https://tex.z-dn.net/?f=T-W%3D%28m_1%2Bm_2%29a)
where T is the tension in the upper sting and W is the weight of the system. Solving the equation for a we have:
![\begin{gathered} 6.6-(0.3+0.24)(9.8)=(0.3+0.24)a \\ a=\frac{6.6-(0.3+0.24)(9.8)}{(0.3+0.24)} \\ a=2.42 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%206.6-%280.3%2B0.24%29%289.8%29%3D%280.3%2B0.24%29a%20%5C%5C%20a%3D%5Cfrac%7B6.6-%280.3%2B0.24%29%289.8%29%7D%7B%280.3%2B0.24%29%7D%20%5C%5C%20a%3D2.42%20%5Cend%7Bgathered%7D)
Therefore the acceleration of the system is 2.42 meters per second per second.
Part B.
Now, that we have the acceleration of the system we analyze the lower block individually; for this block the equation of motion is:
![T^{\prime}-W^{\prime}=m_2a](https://tex.z-dn.net/?f=T%5E%7B%5Cprime%7D-W%5E%7B%5Cprime%7D%3Dm_2a)
where T' is the tension in the lower rope, W' is the weight of the lower block and m2 is its mass. Solving for the tension we have that:
![\begin{gathered} T^{\prime}=(0.24)(9.8)+(0.24)(2.42) \\ T^{\prime}=2.93 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%5E%7B%5Cprime%7D%3D%280.24%29%289.8%29%2B%280.24%29%282.42%29%20%5C%5C%20T%5E%7B%5Cprime%7D%3D2.93%20%5Cend%7Bgathered%7D)
Therefore the tension in the lower rope is 2.93 N