Energy consumed in doing the work = 300 Joules
Force applied on the object = 75 N
Let the distance moved by the object be d.
Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.
Work done = Force x displacement
300 = 75 x d
d = 4 m
Hence, the maximum distance moved by the object = 4 meters
Answer:
A) F(r) = [12a/(r^(13))] - [6b/(r^(7))]
ii) Graphs are attached
B) Equilibrium Distance = (2a/b)^(1/6)
C) Minimum Energy = b²/4a
D) a = 6.67 x 10^(-138) Jm^(12)
b = 6.41 x 10^(-78) Jm^(6)
Explanation:
I've attached the explanation of A-C alongside the graphs
D) i) From the question, we are to make r equal to the derivative of "r" we got when F(r) = 0 which was r = (2a/b)^(1/6)
Thus, since we are given equilibrium distance as: 1.13 x 10^(-10), hence;
(2a/b)^(1/6) = 1.13 x 10^(-10)m
So, (2a/b)= [1.13 x 10^(-10)]^(6)
a = (b/2)[1.13 x 10^(-10)]^(6)
From earlier, we saw that b²/4a = U(r)
Thus since U(r) = 1.54 x 10^(-18) from the question, b²/4a = 1.54 x 10^(-18)
Putting a = (b/2)[1.13 x 10^(-10)]^(6);
We have;
(b²) / [(4b/2)[1.13 x 10^(-10)]^(6)]] = 1.54 x 10^(-18)
b/2 = [1.13 x 10^(-10)]^(6)] x 1.54 x 10^(-18)
So, b = 6.41 x 10^(-78) Jm^(6)
ii) Putting (6.41 x 10^(-78))² for b in;
a = (b/2)[1.13 x 10^(-10)]^(6)
We have, a = (6.41 x 10^(-78))²/ ( 4 x 1.54 x 10^(-18)
So a = 6.67 x 10^(-138) Jm^(12)
