A body of mass m= 1 kg is moving along the x-axis. Its potential energy is given by the function
U(x)=2(x2−1)2
Note: The units were dropped for the numbers in the equation above. You should note that 2 would carry units of J⋅m−4 and 1 would carry units of m2
a) What is the x -component of the force associated with the potential energy given by U(x) ? Give an expression in terms of x.
Fx(x)=
b) At what positive value of x ( x>0) in m, does the potential have a stable equilibrium point ?
x=
c) Suppose the body starts with zero speed at x=1.5 m. What is its speed (in m/s) at x=0 m and at x=−1 m?
v(x=0)=
v(x=−1)=
a) Fx(x)= -8*x*(x^2-1)
b). x=1
c) v(x=-1)= sqrt(6)
c) v(x=0)= 1.5
To find the x-component of the force associated with the potential energy given by U(x), we need to take the derivative of U(x) with respect to x. Let's solve each part of the question step by step:
a) To find Fx(x), we need to take the derivative of U(x) with respect to x:
Fx(x) = dU(x)/dx
First, let's expand U(x):
U(x) = 2(x^2 - 1)^2
= 2(x^4 - 2x^2 + 1)
Now, we can differentiate U(x) with respect to x:
dU(x)/dx = 2(4x^3 - 4x)
Therefore, the x-component of the force associated with the potential energy is:
Fx(x) = 8x^3 - 8x
b) A stable equilibrium point occurs when the force is equal to zero. To find the positive value of x where the potential has a stable equilibrium point, we need to solve the equation:
Fx(x) = 0
8x^3 - 8x = 0
Factoring out 8x:
8x(x^2 - 1) = 0
This equation has two solutions: x = 0 and x = ±1. However, we are looking for a positive value of x, so the stable equilibrium point occurs at x = 1.
c) To find the speed of the body at different positions, we need to consider the conservation of mechanical energy. At any position, the sum of the kinetic energy and potential energy is equal to the total mechanical energy, which is constant.
At x = 1.5 m, we can find the potential energy, U(x), and kinetic energy, K(x):
U(x = 1.5) = 2((1.5)^2 - 1)^2
= 8.5 J⋅m^(-4)
Since the body starts with zero speed, the total mechanical energy at this point is equal to the potential energy:
E = U(x = 1.5) = 8.5 J⋅m^(-4)
Now, let's find the speed at x = 0 m and x = -1 m:
At x = 0 m:
The potential energy, U(x), is given by:
U(x = 0) = 2(0^2 - 1)^2
= 2 J⋅m^(-4)
To find the kinetic energy, we subtract the potential energy from the total mechanical energy:
K(x = 0) = E - U(x = 0)
= 8.5 J⋅m^(-4) - 2 J⋅m^(-4)
= 6.5 J⋅m^(-4)
The speed, v(x = 0), can be found using the relationship between kinetic energy and speed:
K(x = 0) = (1/2)mv^2
Solving for v:
v(x = 0) = sqrt(2K(x = 0)/m)
= sqrt(2*6.5 J⋅m^(-4)/1 kg)
= sqrt(13) m/s
At x = -1 m:
We follow the same process as above:
U(x = -1) = 2((-1)^2 - 1)^2
= 2 J⋅m^(-4)
K(x = -1) = E - U(x = -1)
= 8.5 J⋅m^(-4) - 2 J⋅m^(-4)
= 6.5 J⋅m^(-4)
v(x = -1) = sqrt(2K(x = -1)/m)
= sqrt(2*6.5 J⋅m^(-4)/1 kg)
= sqrt(13) m/s
Therefore, the speed at x = 0 m and x = -1 m is sqrt(13) m/s.