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

Answer:

U=α/(x+x_0)^2

v=1.88m/s

Explanation:

We have that

$F(x)=\frac{\alpha}{(x+x_0)^2}\\\alpha=0.800Nm^2\\x_0=0.200m$

(a)

The potential is calculated by using

$U=-\int Fdx=-\int \frac{\alpha}{(x+x_0)^2}dx\\U=\frac{\alpha}{(x+x_0)}$

(b)

m=0.5kg

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

$F(0.4)=\frac{0.8Nm^2}{(0.4m+0.2m)^2}=2.22N\\F=ma\\2.22N=(0.5kg)a\\a=4.44\frac{m}{s^2}$

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

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

I hope this is useful for you

regards