The figure shows a plot of potential energy U versus position x of a 0.260 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.260 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 6.00 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side?

Explanation / Answer

Here we will apply the conservation of energy
(a) At X = 3.5 m
Total energy = UA + (K.E)A
At point B it is given that
Total energy at B = UB +KB = 12+6 = 18 J
so total energy at A = total energy at B
9+K.EA = 18
K.EA = 9 J
(1/2)mV2 = 9
V = 8.32 m/s where V is velocity and m is mass
(b) Similarly at X = 6.5 m potential energy is zero
so K.E = 18 J
V2 = 18*2 /0.260
V = 11.77 m/s
(c) At point c the potentiaal energy becomes 20 J therefore the total energy will exhausted and velocity becomes zero therefore ball will turn the direction.
Poistion toward right wil be
20 /x = 24
X = 20/24 = 0.83
Toward right = 7+0.83 = 7.83 m
(d)and toward left of d = 1-0.83 = 0.167 m

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