A 0.60-kg particle moves along the x-axis. Its potential energy is given by U(x)=5x^2, where U is joules and x is in meters. If the particle has a speed of 3.0m/s what it is at x=1.2m, its speed when it is at the origin is?
If total energy is constant then
(1/2) m v^ 2 + 5 x^2 = constant = c
.3 v^2 + 5 x^2 = c
.3 (3)^2 + 5 (1.2)^2 = c
solve for c
then put in x = 0
.3 v^2 = c
To find the speed of the particle when it is at the origin (x=0), we need to use the conservation of mechanical energy.
The total mechanical energy (E) of the particle is the sum of its kinetic energy (K) and potential energy (U). According to the conservation of mechanical energy, the total mechanical energy remains constant throughout the motion:
E = K + U
Given that the potential energy function is U(x) = 5x^2 and the particle's speed is 3.0 m/s at x = 1.2 m, we can first find the total mechanical energy (E) at x = 1.2 m:
E = K + U
E = 0.5 * m * v^2 + 5 * x^2
E = 0.5 * 0.60 kg * (3.0 m/s)^2 + 5 * (1.2 m)^2
Calculating the equation above will give us the value of E.
Next, we can find the kinetic energy (K) at the origin by setting x = 0:
E = K + U
E = 0.5 * m * v_0^2 + 5 * (0 m)^2
Since the potential energy at the origin is zero (U = 0), the equation simplifies to:
E = 0.5 * m * v_0^2
We can now equate the two expressions for E to find the kinetic energy (K) at the origin:
0.5 * 0.60 kg * (3.0 m/s)^2 + 5 * (1.2 m)^2 = 0.5 * 0.60 kg * v_0^2
Solving for v_0, the speed at the origin, will give us the answer.