Answer:
The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(C) is correct option.
Explanation:
Given that,
Mass of block = 1.0 kg
Dependent force = F(x)
Frictional force = F(f)
Suppose, the following information would students need to test the hypothesis,
(A) The function F(x) for 0 < x < 5 and the value of F(f).
(B) The function a(t) for the time interval of travel and the value of F(f).
(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).
(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).
We know that,
The work done by a force is given by,
.....(I)
Where,
= net force
We know, the net force is the sum of forces.
So, ![\sum{F}=ma](https://tex.z-dn.net/?f=%5Csum%7BF%7D%3Dma)
According to question,
We have two forces F(x) and F(f)
So, the sum of these forces are
![F(x)+(-F(f))=ma](https://tex.z-dn.net/?f=F%28x%29%2B%28-F%28f%29%29%3Dma)
Here, frictional force is negative because F(f) acts against the F(x)
Now put the value in equation (I)
![W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}](https://tex.z-dn.net/?f=W%3D%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7B%28F%28x%29-F%28f%29%29dx%7D)
We need to find the value of ![\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}](https://tex.z-dn.net/?f=%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7B%28F%28x%29-F%28f%29%29dx%7D)
Using newton's second law
...(II)
We know that,
Acceleration is rate of change of velocity.
![a=\dfrac{dv}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D)
Put the value of a in equation (II)
![\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}](https://tex.z-dn.net/?f=%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7B%28F%28x%29-F%28f%29%29dx%7D%3D%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7Bm%5Cdfrac%7Bdv%7D%7Bdt%7Ddx%7D)
![\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}](https://tex.z-dn.net/?f=%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7B%28F%28x%29-F%28f%29%29dx%7D%3D%5Cint_%7Bv_%7B0%7D%7D%5E%7Bv_%7Bf%7D%7D%7Bmv%5C%20dv%7D)
![\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}](https://tex.z-dn.net/?f=%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx_%7Bf%7D%7D%7B%28F%28x%29-F%28f%29%29dx%7D%3D%5Cdfrac%7Bmv_%7Bf%7D%5E2%7D%7B2%7D%2B%5Cdfrac%7Bmv_%7B0%7D%5E2%7D%7B2%7D)
Now, the work done by the net force on the block is,
![W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7Bmv_%7Bf%7D%5E2%7D%7B2%7D%2B%5Cdfrac%7Bmv_%7B0%7D%5E2%7D%7B2%7D)
The work done by the net force on the block is equal to the change in kinetic energy of the block.
Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(C) is correct option.