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.

Given values are:

As we know,

→

→

Thus the response above is right.

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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

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