![](https://tex.z-dn.net/?f=%3Cb%3E%3Ci%3E)
Hi pupil Here's Your answer :::
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Student's justification is not correct. Two equal and opposite force cancel each other if the act on the same body. According to the third law of motion action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.
When we push a message track, then the applied force on the truck is not sufficient to overcome the force of friction between the tyres of truck and ground, hence, truck does not move.
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Hope this helps .
Force can be exerted into an object with out it moving, but if you were to move the object due to force it would be considered work. (yes, you can exert force without having the object moving)
The difference between a fuse and a circuit breaker is that a fuse is a one time circuit element while a circuit breaker is a multiple use device. If you were designing a circuit for a reading lamp, you should include a circuit breaker so you don't have to replace it every time.
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.