It B as i think it fits in the substrate of the atoms
Answer:
![k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B2%5Ccdot%20m%20%5Ccdot%20g%20%5Ccdot%20%28d%2Bx_%7Bf%7D%29%5Ccdot%20%28%5Csin%20%5Ctheta%20-%20%5Cmu_%7Bk%7D%5Ccdot%20%5Ccos%20%5Ctheta%29%7D%7Bx_%7Bf%7D%5E%7B2%7D%7D)
Explanation:
Let assume that spring reaches its maximum compression at a height of zero. The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:
![U_{g,A}=U_{k,B} + W_{f}](https://tex.z-dn.net/?f=U_%7Bg%2CA%7D%3DU_%7Bk%2CB%7D%20%2B%20W_%7Bf%7D)
![m\cdot g \cdot (d + x_{f})\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x_{f}^{2}+\mu_{k}\cdot m \cdot g \cdot (d+x_{f})\cdot \cos \theta](https://tex.z-dn.net/?f=m%5Ccdot%20g%20%5Ccdot%20%28d%20%2B%20x_%7Bf%7D%29%5Ccdot%20%5Csin%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20k%20%5Ccdot%20x_%7Bf%7D%5E%7B2%7D%2B%5Cmu_%7Bk%7D%5Ccdot%20m%20%5Ccdot%20g%20%5Ccdot%20%28d%2Bx_%7Bf%7D%29%5Ccdot%20%5Ccos%20%5Ctheta)
![m\cdot g \cdot (d + x_{f})\cdot (\sin \theta-\mu_{k}\cdot \cos \theta) = \frac{1}{2}\cdot k \cdot x_{f}^{2}](https://tex.z-dn.net/?f=m%5Ccdot%20g%20%5Ccdot%20%28d%20%2B%20x_%7Bf%7D%29%5Ccdot%20%28%5Csin%20%5Ctheta-%5Cmu_%7Bk%7D%5Ccdot%20%5Ccos%20%5Ctheta%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20k%20%5Ccdot%20x_%7Bf%7D%5E%7B2%7D)
The spring constant is cleared in the expression described above:
![k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B2%5Ccdot%20m%20%5Ccdot%20g%20%5Ccdot%20%28d%2Bx_%7Bf%7D%29%5Ccdot%20%28%5Csin%20%5Ctheta%20-%20%5Cmu_%7Bk%7D%5Ccdot%20%5Ccos%20%5Ctheta%29%7D%7Bx_%7Bf%7D%5E%7B2%7D%7D)
The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057
Answer:
a) 4.6*10^{-5} J
b) 6133.33 J/C
c) 94358.9 C
Explanation:
(a) We have that the change in the kinetic energy equals the net work over the charge. Hence we have
![\Delta E_K=W-W_E](https://tex.z-dn.net/?f=%5CDelta%20E_K%3DW-W_E)
where Ek is the kinetic energy, W is the work of the external force and WE is the work done by the electric field. By replacing we obtain:
![W_E=W-\Delta E_k=7.60*10^{-5}J-3*10^{-5}J=4.6*10^{-5}J](https://tex.z-dn.net/?f=W_E%3DW-%5CDelta%20E_k%3D7.60%2A10%5E%7B-5%7DJ-3%2A10%5E%7B-5%7DJ%3D4.6%2A10%5E%7B-5%7DJ)
(b) The potential difference is computed by using:
![\Delta V=\frac{W_E}{q}=\frac{4.6*10^{-5}J}{7.5*10^{-9}C}=6133.33\frac{J}{C}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cfrac%7BW_E%7D%7Bq%7D%3D%5Cfrac%7B4.6%2A10%5E%7B-5%7DJ%7D%7B7.5%2A10%5E%7B-9%7DC%7D%3D6133.33%5Cfrac%7BJ%7D%7BC%7D)
(c) With the work done by the electric force we can calculate the Electric field. By using the following formula we obtain:
![W_E=qEd\\\\E=\frac{W_E}{qd}=\frac{4.6*10^{-5}J}{(7.5*10^{-9}C)(6.50*10^{-2}m)}=94358.9\frac{N}{C}](https://tex.z-dn.net/?f=W_E%3DqEd%5C%5C%5C%5CE%3D%5Cfrac%7BW_E%7D%7Bqd%7D%3D%5Cfrac%7B4.6%2A10%5E%7B-5%7DJ%7D%7B%287.5%2A10%5E%7B-9%7DC%29%286.50%2A10%5E%7B-2%7Dm%29%7D%3D94358.9%5Cfrac%7BN%7D%7BC%7D)
where we have used d=6.5cm=6.5*10^-9m and q=7.5nC=7.5*10^-9C
hope this helps!!