Question

a single conservative force act on body of mass 1kg that moves along the x-axis the potential energy v(x) is given by v(x)=20+(x-2)^2 where x is in the meters . at x=5.0m the particle has a kinetic energy 20j then find
a) minimum value of x in metres
b)maximum value of x in metres
c)maximum potential energy in joules
d)maximum kinetic energy in joules

Answer 1

Answer:

what is the correct answer of this question?

Answer 2

The maximum value of x in meters is 7.38 m and minimum value is  − 3.38 m

Explanation:

We are given that:

• Mass of body = 1 Kg
• Distance "x" = 5.0 m
• Kinetic energy of particle K.E = 20 J

The greatest or least value of x will be at the point where Kinetic energy is zero and U is maximum.

U(max) = 49 J

⇒ 20 + (x − 2) 2 = 49

⇒ (x − 2)2 = 29

⇒ x − 2 = ±√29

⇒ x = −3.38 m, 7.38 m

Thus the maximum value of x in meters is 7.38 m and minimum value is  −3.38 m