Question

vii) Find the percentage error in kinetic
energy of a body having mass 60.0 +
0.3 g moving with a velocity 25.0 + 0.1
cm/s. ​

Answer 1

Explanation:

kinetic energy is given by, K = 1/2 mv²

where K is kinetic energy, m is mass of body and v is velocity of body.

to find error in kinetic energy, let's first derive formula of % error in kinetic energy.

K = 1/2 mv²

taking log both sides,

logK = log(1/2mv²)

logK = log(1/2) + logm + 2logv

now differentiate both sides ,

dK/K = 0 + dm/m + 2dv/v

or, ∆K/K × 100 = ∆m/m × 100 + 2∆v/v × 100

hence, % error in kinetic energy = ∆m/m × 100 + 2∆v/v × 100

given, mass = (60 ± 0.3)g

so, ∆m = 0.3 and m = 60

similarly, velocity = (25 ± 0.1)

so, ∆v = 0.1 and v = 25

then, % error in kinetic energy = (0.3/60) × 100 + 2 × (0.1/25) × 100

= 1/2 + 4/5

= (5 + 8)/10

= 1.3 %