A single conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. Thepotential energy
U(x) is given by: U(x) = 20 + (x – 2)
2 where x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J.(i) What is the mechanical energy of the system?(ii) Make a plot of U (x) as a function of x for – 10 m < x < 10m, and on the same graph draw the line that represents the mechanical energy of the system. Use part (ii) to determine(iii) The least value of x and(iv) The greatest value of x between which the particle can move.(v) The maximum kinetic energy of the particle and(vi) The value of x at which it occurs.(vii) Determine the equation for F (x) as a function of x.(viii) For what value of x does F(x) = 0?
Text Solution
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(i) 49 J
(ii) 
(iii) 
≈ –3.38 m
(iv) 
≈ 7.38 m
(v) 29 J
(vi) x = 2m
(vii) F = 2 (2 – x)
(viii) x = 2
Sol. U (x) = 20 + (x – 2) 2
= 2(x – 2)
– F = 2(x – 2)
F = – 2(x – 2)
m (x – 2) = – 2 (x – 2)
Let x = x – 2
mx = – 2 x
1 x = – 2 x
x = – 2 x Simple Harmonic Motion
Mean position is x = x – 2 = 0 ⇒ x = 2
W 2 = 2,
Kinetic energy = 
mv 2
= 
(1) ( ω 2 ) (A 2 – x 2 ) = x – 2, x = 5 – 2 = 3
20 = 
(1) (2) {A 2 – 3 2 }
20 = A 2 – 9 ⇒ A 2 = 29 ⇒ A = 
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