Your answer is partially correct. The figure shows a plot of potential energy U

Question

Your answer is partially correct. The figure shows a plot of potential energy U versus position x of a 0.250 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA 9.00 J,Uc- 20.0 J and Up 24.0 J. The particle is released at the point where U forms a "potential hill" of "height Ug 12.0 J, with kinetic energy 5.00 J. What is the speed of the particle at (a)x 3.50 m and (bx 6.50 m? What is the position of the turning point on (c) the right side and (d) the left side? x (m) ) Number 12.3 Units m/s ) Number Units 11.4 m/s ) Number Units 45.6 ) Number Units 3.00

Explanation / Answer

It requires law of conservation of energy

a)

P1 + K1 = P2 + K2

12 + 5 = 9 + K2 because PE = 9 J at x = 3.5m

K2 = 8 J

but K2 = 1/2 mv^2

8 = 1/2 x 0.25 x v^2

v = 8 m/s

b)

P1 + K1 = P2 + K2

12 + 5 = 0 + K2 because PE = 0 J at x = 6.5m

K2 = 17 J

but K2 = 1/2 mv^2

17 = 1/2 x 0.25 x v^2

v = 11.66 m/s

c)

From graph, it can be seen at 7m

d)

From graph, it can be seen at around 5m.

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