A single conservative force Fx= (6.0x-12) N, (x is in m), acts on a particle moving along the x axis. The potential energy associated with this force is assigned a value of 20J at x=0. What is the potential energy at x=3.0m?
Oh, how calculus explains it all..
the average force going the three meters is (6*3-12)-(6*0-12)/2
so the work put in to the object during pushing is
avgforce*distance=6*3/2 * 3=27 joules
so the PE at the end is 27+20J
check my math.
To find the potential energy at x = 3.0 m, we need to determine the work done by the conservative force over that displacement. The work done by a conservative force is equal to the negative change in potential energy.
In this case, the potential energy associated with the force is already given as 20 J at x = 0. Therefore, we can define the potential energy function as:
U(x) = -integral of F(x) dx,
where F(x) is the force function and U(x) is the potential energy function.
Given that F(x) = 6.0x - 12 N, we can integrate this function over the displacement from x = 0 to x = 3.0 m.
So, to find the potential energy at x = 3.0 m, we need to calculate the following integral:
U(x) = -integral of (6.0x - 12) dx,
Let's break down the integral and solve it step by step:
U(x) = -integral of 6.0x dx + integral of (-12) dx,
Integrating each term,
U(x) = -[(6.0/2)x^2 + (-12)x] + C,
U(x) = -3.0x^2 + 12x + C,
Now, we need to determine the constant of integration, C.
Given that the potential energy is assigned a value of 20 J at x = 0, we can substitute these values into our potential energy function:
20 = -3.0(0)^2 + 12(0) + C,
20 = C.
Therefore, the constant of integration is 20.
Now, we can substitute x = 3.0 m into our potential energy function:
U(x) = -3.0(3.0)^2 + 12(3.0) + 20,
U(x) = -27 + 36 + 20,
U(x) = 29 J.
Hence, the potential energy at x = 3.0 m is 29 J.