You would have to run a little less than 2 blocks
Answer:

Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

The initial potential energy of the spring is given by the equation:

the Kinetic energy of the block is then given by the equation:

so we can now set them both equal to each other, so we get:

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

so now we can solve this for the final velocity, so we get:

The best answer to go with is b
Answer:
False.
Explanation:
The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,
FBA = −FAB
However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:
atruck = F/M
acar = F/m
and combining these we get:
atruck/acar = m/M
So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of
(Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

![W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5Cbig%20%5B15y-%5Cfrac%7By%5E2%7D%7B2%7D%5Cbig%20%5D%5E%7B15%7D_0)
![W = (14*7*62)[15(15)-\frac{(15)^2}{2}]](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5B15%2815%29-%5Cfrac%7B%2815%29%5E2%7D%7B2%7D%5D)

Therefore the total work in the system is 