If an object with a mass of #15 kg # changes speed from #8 m/s# to #14 m/s#, by how much does its kinetic energy change?

Answer 1

I found a change of #DeltaK=990J#.

Kinetic Energy #K# is given as: #K=1/2mv^2# Considering the initial and final stages of your phenomenon you have that the change in #K# will be: #DeltaK=K_f-K_i=1/2mv_f^2-1/2mv_i^2=1/2m(v_f^2-v_i^2)=# #=1/2*15*(14^2-8^2)=990J#

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Answer 2

Adam Arnold

The change in kinetic energy can be calculated using the formula ΔKE = KE_final - KE_initial. KE = 0.5 * mass * velocity^2. Substituting the values, ΔKE = 0.5 * 15 kg * (14 m/s)^2 - 0.5 * 15 kg * (8 m/s)^2. Calculating, ΔKE = 0.5 * 15 kg * 196 m^2/s^2 - 0.5 * 15 kg * 64 m^2/s^2 = 0.5 * 15 kg * 132 m^2/s^2 = 990 J.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.