Question

A pendulum of mass 2.0 kg is raised to a height of 0.4 m above the lowest point in its swing and then is released from rest. If air resistance can be ignored, how high will the pendulum swing on the other side of its motion? A) Half as high B) One fourth as high C) One third as high D) Just as high E) Not move 9) 3pt For the pendulum in the previous problem, how fast will it move at the lowest point in its swing? Show all work or no credit A) 0.5 m/s B) 1 m/s C) 5m/s D) 10 m/s E) none of these Q10) 3pt A spring of spring constant 60 N/m is stretched a distance of 0.3 m from its equilibrium position. Calculate the increase in the potential energy of the spring. Show all work or no credit. A)45 joules B) 9 joules C) 8 joules D) 2.0 joules E) none of these show all work with steps and equations

Answer 1

1) (D) Just as high.

A pendulum swing is simple harmonic motion, so the pendulum is at rest at the extremes of its motion, and it possesses only potential energy at those positions. Thus the potential energy at each extreme must be the same. The potential energy of a pendulum depends only upon the height of the pendulum above the lowest point, so the two heights must be the same. Thus the answer is 0.4 m.

2) (E) none of these.

we have

PE1 + KE1 = PE2 + KE2

m g h1 + 0 = 0 + 1 / 2 m (v2)^2

g h1 = 1 / 2 (v2)^2

(9.8 m / s2) (0.4 m) = (1 / 2) (v2)^2

(v2)^2

= 7.84 m2 / s2

v2 = 2.8 m / s

3) (E) None of these.

PEs = 1 / 2 k x^2

PEs = (1 / 2) (60 N / m) (0.3 m^)2

PEs = 2.7 N m = 2.7 J