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.
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Answered by
170
answer : 1.3%
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 %
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