Question

A one-dimensional conservative force is given by the function F(x) = (2.00 N/m) x +(1.00 N/m3) x3. What is the change in potential energy of the object as it moves from x = 1.00 m to x = 2.00 m? Enter your answer in Joules.

Answer 1

Answer:

Explanation:

Given that

F(x) = (2.00 N/m) x +(1.00 N/m3) x3.

Which can be written as

F(x) = 2x +x³ N/m

Potential energy from x=1 to x=2

P.E is given as

P.E=∫F(x)dx

P.E=∫2x+x³ dx. From x=1 to x=2

P.E= 2x²/2 + x⁴/4 from x=1 to x=2

P.E= x² + x⁴/4 from x=1 to x=2

P.E=(2²+2⁴/4)-(1² +1⁴/4)

P.E=(4+4)-(1+1/4)

P.E, =8-5/4

P.E=27/4

P.E=6.75J

Since it is moving from lower position to upper position, then P.E should be negative

P.E = - 6.75J

Answer 2

The change in Potential energy will be "-6.75 Joule".

Given values are:

• $F(x) = (2x+1x^3) N$

As we know,

→ $\Delta U = - \int_{x_1}^{x_2} F.dx$

→ $\Delta U = - \int_1^2 (2x+x^3)dx$

         $= -[x^2+\frac{x^4}{4} ]_1^2$

         $= -(8-\frac{5}{4} )$

         $= -\frac{27}{4}$

         $= -6.75 \ Joule$

Thus the response above is right.

Learn more about P.E here:

https://brainly.com/question/13898544