Question

A ball of mass 0.16kg is thrown upwards with an initial velocity of 25ms-1 and reaches a maximum height of 20m. Calculate the percentage loss of energy due to air resistance.

Answer 1

A ball of mass 0.16kg is thrown upwards with an initial velocity of 25ms-1 and reaches a maximum height of 20m.The percentage loss of energy due to air resistance is 36%

Given :

• Mass of ball =0.16 kg

• Initial velocity of the ball =25m/s.

• Maximum height achieved =20 m

To find :

• Percentage loss of energy due to air resistance .

Solution :

• Let there is no air resistance

According to law of conservation of energy

"the total energy of an isolated system remains constant"

KE of body will be converted into PE

ΔK.E.=ΔP.E.

$Kinetic \: energy = \frac{1}{2} \times m \times {v}^{2}$

• As at highest point final velocity (v) will be 0

To calculate maximum height achieved

• ΔK.E=ΔP.E

• ½m(v²-u²)=mg(h2-h1)

Solving equation -

• ½m(0-25²)=m×g×h (taking g=10m/s²)

• $h = \frac{1}{2} \times \frac{625}{10} = 31.25m$
• If no air resistance is there particle would have gone 31.25 m

• But Particle achieved only 20m

Total energy of particle=½mu²

=½×0.16×625=50 J

• Loss in energy = mg(h2-h1)

=0.16×10×(31.25-20)

=1.6×11=18J

• Loss in energy due to air resistance =18J

To find % energy loss .

• $Energy \: loss = \frac{ \:Loss \: of energy}{ \:total energy} ×100</u></strong></li><li><strong><u>[tex] Energy \: loss = \frac{ \:Loss \: of energy}{ \:total energy} ×100$
• Energy loss = 18/50×100

=36%

Loss of energy due to air resistance =36%