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If an object with a mass of # 3kg # changes speed from #5m/s# to # 1m/s#, by how much does its kinetic energy change?

Answer 1

Nathan Ashcroft

#Delta E_k=-36 J#

#v_f=1 m/s# #v_i=5 m/s# #m=3 kg#

#Delta E_k=1/2*m(v_f^2-v_i^2)#

#Delta E_k=1/2*3*(1^2-5^2)#

#Delta E_k=3/2(1-25)#

#Delta E_k=-3/cancel(2)*cancel(24)^12#

#Delta E_k=-3*12#

#Delta E_k=-36 J#

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Answer 2

Lily Adams

The change in kinetic energy can be calculated using the formula:

ΔKE = ½ * m * (v₂² - v₁²)

Where:

ΔKE is the change in kinetic energy,
m is the mass of the object (3kg),
v₁ is the initial velocity (5m/s),
v₂ is the final velocity (1m/s).

Plugging in the values:

ΔKE = ½ * 3kg * ((1m/s)² - (5m/s)²) ΔKE = ½ * 3kg * (1 - 25) ΔKE = ½ * 3kg * (-24) ΔKE = -36 J

Therefore, the kinetic energy decreases by 36 joules.

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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.