A 0.20-kg particle moves along the x axis under the influence of a stationary ob

Question

A 0.20-kg particle moves along the x axis under the influence of a stationary object. The potential energy in SI unit is given by U(x) = 80.0x + 20.0x^2 where x is the position of the particle.  

a) If the particle has a speed of 5.0 m/s when it is at x = 2.0 m, what was its speed at the origin.             
b) If the total mechanical energy is 60.0 J, find the turning points of the motion.                   
c) When the particle is at x = 1.0m find the magnitude and direction of its acceleration

Explanation / Answer

a)we know that

v^2 - u^2 = 2gS

or u^2 = v^2 - 2gS

or u = (v^2 - 2gS)^1/2

where v = 5 m/s,g = 9.8 m/s^2 and S = 2.0 m

b)total energy is U = 60.0J

or 60.0 = 80.0x + 20.0x^2

or 20.0x^2 + 80.0x - 60.0 = 0

the roots are x1 = -4.64 m and x2 = 0.6457 m

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