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

Answer 2

The potential energy function for this force will be: $U=\frac{a}{X+X_0}$

The speed that the object will reach will be equal to 1.88m/s.

We can arrive at this answer because:

• To find the potential energy function, we must look at the root of the calculation of potential energy. This is done as follows:

$U=- \int\limits F dx\\U= - \int\limits\frac{a}{(x+x_0)^2} * dx\\U= \frac{a}{x=x_0}$

• To find the object's velocity, we have to first find its acceleration and for that, we will use the force of x equal to 4. Therefore, we can do the following calculation:

$F(0.4)= \frac{0.8Nm^2}{(0.4m+2.4m)^2}= 2.22N\\\\\\F= m*a\\2.22= (0.5)*a\\a=\frac{2.22}{0.5} \\a= 4.44 m/s^2$

• From that value, we can calculate the speed. This will be done as follows:

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

More information:

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