Question

A conservative force F⃗ is in the +x-direction and has magnitude F(x)=α/(x+x0)2, where α=0.800N.m2 and x0=0.200m. (a) What is the potential energy function U(x) for this force? Let U(x)→0 as x→ω (b) An object with mass m = 0.500 kg is released from rest at x = 0 and moves in the +x-direction. If F⃗ is the only force acting on the object, what is the object’s speed when it reaches x = 0.400 m?

Answer 1

The potential energy function for the given conservative force is U(x) = - α/(x+x0) + α/x0. The constant of integration is determined using the condition that U(x) → 0 as x → ∞, which gives C = α/x0. The final potential energy function is U(x) = - α/(x+x0) + α/x0.

Step-by-step explanation:

To find the potential energy function for the conservative force F⃗ , we need to integrate the force function with respect to x. The potential energy function U(x) is given by U(x) = - ∫ F(x) dx + C, where C is the constant of integration. In this case, F(x) = α/(x+x0)², so we obtain U(x) = - α/(x+x0) + C.

To determine the constant of integration, we use the condition that U(x) → 0 as x → ∞. This implies that the potential energy at infinite separation is zero, so U(∞) = 0. Substituting this into the U(x) equation and solving for C, we find C = α/x0.

Therefore, the final potential energy function is U(x) = - α/(x+x0) + α/x0, where α = 0.800 N·m² and x0 = 0.200 m.

Answer 2

U=α/(x+x_0)^2

v=1.88m/s

Step-by-step explanation:

We have that

`F(x)=(\alpha)/((x+x_0)^2)\\\alpha=0.800Nm^2\\x_0=0.200m`

(a)

The potential is calculated by using

`U=-\int Fdx=-\int (\alpha)/((x+x_0)^2)dx\\U=(\alpha)/((x+x_0))`

(b)

m=0.5kg

The acceleration can be obtained if we calculate the force for x=4, and after we compute the acceleration

`F(0.4)=(0.8Nm^2)/((0.4m+0.2m)^2)=2.22N\\F=ma\\2.22N=(0.5kg)a\\a=4.44(m)/(s^2)`

and finally, we can use the equation for the final speed

`v^2=v_0^2+2ax\\\\v=\sqrt{(0)+2(4.44(m)/(s^2))(0.4m)}\\\\v=1.88(m)/(s)`

I hope this is useful for you

regards