Question

The percentage error in the mass and speed are 2% and 3% respectively. What is the the maximum error in kinetic energy calculated using these quantities?

Answer 1

Given :  

Δm/m x 100= 2%

Δv/v x100= 3%

K.E=K=  1/2 mv²

(ΔK/K) x100=(Δm/m )x100+ 2 (Δv/v) x100

=2+2(3)

=2+6

=8%

∴The maximum error in kinetic energy IS 8%

Answer 2

Hii mate,

# Answer- 8 %.

# Given-
Δm/m = 2 %
Δv/v = 3 %

# Explaination-
Let's solve this step-by-step.

Formula for kinetic energy is,
K.E. = (1/2)mv^2

Percentage error in kinetic energy will be,
ΔK.E./K.E.×100 = Δm/m×100 + 2Δv/v×100
Δd/d×100 = 2 % + 2×3 %
Δd/d×100 = 8 %.

Percentage error in kinetic energy is 8 %.

Hope that was useful...