They stay with the microscope as it moves around to different schools, and they are always located in the same classroom where the rest of the microscope is being used.
The answer is 60 km. I hope it helps i dont know if this is right or wrong.
Answer:
a) ![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
b) ![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
c) ![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
Part b
For this case first we can convert the spring constant to N/m like this:
![2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}](https://tex.z-dn.net/?f=2%20%5Cfrac%7BN%7D%7Bcm%7D%20%5Cfrac%7B100cm%7D%7B1m%7D%3D200%20%5Cfrac%7BN%7D%7Bm%7D)
And the work donde by the spring on this case is given by:
![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
![W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i](https://tex.z-dn.net/?f=%20W_%7Bg%7D%20%2BW_%7Bspring%7D%20%3D%20K_%7Bf%7D%20-K_%7Bi%7D%3D0-%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2_i)
And if we solve for the initial velocity we got:
![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
![-1/2mV^2_i = W_g + W_{spring}](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3D%20W_g%20%2B%20W_%7Bspring%7D)
And replacing we got:
![-1/2mV^2_i =mg d_1 -1/2 k d^2_1](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3Dmg%20d_1%20-1%2F2%20k%20d%5E2_1)
And we can put the terms like this:
![\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2_1%20-mg%20d_1%20-1%2F2%20m%20V%5E2_i%20%3D0)
If we multiply all the equation by 2 we got:
![k d^2_1 -2 mg d_1 -m V^2_i =0](https://tex.z-dn.net/?f=%20k%20d%5E2_1%20-2%20mg%20d_1%20-m%20V%5E2_i%20%3D0)
Now we can replace the values and we got:
![200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0](https://tex.z-dn.net/?f=200N%2Fm%20d%5E2_1%20-0.21kg%289.8m%2Fs%5E2%29d_1%20-0.21%20kg%285.50%20m%2Fs%29%5E2%29%20%3D0)
![200 d^2_1 -2.058 d_1 -6.3525=0](https://tex.z-dn.net/?f=200%20d%5E2_1%20-2.058%20d_1%20-6.3525%3D0)
And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.