A student that is running in a gym at a speed of 3.5m/s grabs the rope hanging from the ceiling and swings on it. a. how high will he swing? [63cm]
b. How high will he be when his speed reduced to half of its initial value? [16cm, ¼ of the initial value]

Can someone explain the logic behind the second part of the question (why is it 1/4 the initial value)?

a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,

K = 1/2 m (3.5 m/s)²

and at the highest point of the swing, the student only has potential energy

P = mgh

The energies at the bottom and top of the swing must be equal, so

1/2 m (3.5 m/s)² = mgh

h = (3.5 m/s)² / (2g)

h = 0.625 m ≈ 63 cm

b. In part (a), we found the relationship

h = v²/(2g)

If we cut the speed v in half, we replace v in the equation above with v/2 :

h = (v/2)²/(2g)

and simplifying this gives

h = (v²/4)/(2g) = 1/4 • v²/(2g)

The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as

1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j