Respuesta :
Answer: correct option is C
The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Ff.
Step-by-step explanation:
The total work done = P.E + K.E
K.E = 1/2mv^2
The speed V depends on the position:
0 < x < 5
Also, since work done = F × distance x
The workdone also depends on F(x)
The net force = 0 that is
F + fr = 0
Where fr = frictional force.
Therefore the best option for the hypothesis of work done will be;
The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Ff.
Answer:
(C) The function F(x) for 0<x<5, the block's initial velocity, and the value of Ff
Step-by-step explanation:
From physics, we know that the work done by a force is given by the following integral:
[tex]Work=\int\limits^{x_f}_{x_0} {F(x)} \, dx [/tex]
Where F(x) is the net force function.
The net force is given by the sum of forces:
[tex]\Sigma F=ma[/tex]
In this case there are only two forces we care about, F(x) and Ff, so the sum of forces will be:
[tex]F(x)-Ff=ma[/tex]
In this case Ff is negative because it acts against the force F(x)
So our integral will now look like this:
[tex]\int\limits^{x_f}_{x_0} {F(x)-Ff} \, dx = \int\limits^{x_f}_{x_0} {ma} \, dx [/tex]
Now, we also know that the acceleration is defined to be:
[tex]a=\frac{dv}{dt}[/tex]
so we can do the substitution:
[tex]\int\limits^{x_f}_{x_0} {F(x)-Ff} \, dx = \int\limits^{x_f}_{x_0} {m\frac{dv}{dt}} \, dx [/tex]
we also know that:
[tex]\frac{dx}{dt}=v[/tex]
so we can substitute again and change the limits of integration of the right side of the equation, so we get:
[tex]\int\limits^{x_f}_{x_0} {F(x)-Ff} \, dx = \int\limits^{v_f}_{v_0} {mv} \, dv [/tex]
so when solving the integral we get:
[tex][F(x)-Ff](x_f-x_0)=\frac{mv_{f}^{2}}{2}-\frac{mv_{0}^{2}}{2}[/tex]
the right hand side of the equation:
[tex]\frac{mv_{f}^{2}}{2}-\frac{mv_{0}^{2}}{2}[/tex]
will be the difference of kinetic energy which proves the hypothesis.
In this case, we know what the mass is, and the x-values. So besides the final velocity, we need to know the initial velocity, the force F(x) for 0<x<5 and the friction Ff, so the answer will be (C)