As a 3.0 kg object moves from point A to point B, it is acted upon by a single c

Question

As a 3.0 kg object moves from point A to point B, it is acted upon by a single conservative force which does -40 J of work during this motion. At point A the speed of the particle is 6.0 m/s and the potential energy associated with the force is +50 J. What is the potential energy at point B? +14J +12J +90J +23J +68J A 4.0-gram particle is moving horizontally rightward with a speed of 5.0 m/s when it strikes a 9.0-gram particle that is initially at rest The 4.0 gram particle rebounds and, following the collision, moves leftward with a spe

Explanation / Answer

use the law of conservation

work done by force will reduce the kinetic energy

intial kinetic energy= 1/2 m v2 = 3 X 36/2 = 108/2 = 54 J

work done by conservative force = -40 J => loss in kinetic energy = -40 J

final kinetic energy will be = 54 - 40 = 14 J

since only conservative forces are there then

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

54 + 50 = 14 + final potential energy

final potential energy = 104 - 14= 90 J => option (c)

(b) law of conservation of momentum

m1 v1 + m2 v2 = m1 v'1 + m2 v'2

v1 = 5 m/s andv2 = 0 , v'1 = 1.5 m/s and v'2 = we have to find

4 X 5 + 9 X 0 = 4 X 1.5 + 9 X v'2     => v'2 = 1.555 m/s towards right side.

so momentum= m2 v'2 = 9 X 10-3 X 1.555= 0.014 N-s (option (a) )

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