At the low point in its swing, a pendulum bob with a mass of 0.5 kg has a velocity of 4 m/s.?

Question

A. What is its kinetic energy at the low point?

B. Ignoring air resistance, how high will the bob swing above the low point before reversing direction?

Answer 1

A.

KE = (1/2)m v²  = (1/2)(0.5)(4²) = 4 J

B.

Just before reversing direction, all KE of the bob has been conversed into PE

PE = mgh

h = PE/mg = 4 / 0.5 x 9.8 = 0.82 m

Answer 2

A. KE = .5 * M * V²

.5 * .5  * 4² = .25 * 16 = 4.

4joules.

B.

PE = KE = 4joules

PE = mgh

4 = .5 * 9.8 * h

4/(.5 * 9.8) = h = .816

.8meters

Answer 3

So, at the lowest point of the pendulum's swing (the point at which all its energy is kinetic) the bob has a velocity of 4m/s.

So, remember the kinetic energy equation,

T=1/2 m v^2      where T denotes the kinetic energy.

So, T= 1/2 (.5 kg) (4m/s)^2

T=4 J

For the second part of the question,

Remember the equation for potential energy:

U=mgh      where U denotes potential

h=U/(m*g)

If we let all the kinetic energy into potential and assume a gravity of 10 m/s^2,

h=(4J)/(.5kg*10m/s^2)

h=0.8m

hope this helps =)