Respuesta :
Answer:
Explanation:
Case 1:
Vo = 0 m/s
V1 = 15 m/s
Case 2:
Vo = 15 m/s
V1 = 30 m/s
Change in KE = 1/2 × m × (V1^2 - Vo^2)
KE1 = 1/2 × 1000 × (15^2 - 0)
= 112.5 kJ
KE2 = 1/2 × 1000 × (30^2 - 15^2)
= 1/2 × 1000 × 675
= 337.5kJ
Case 1 has an increase of 112.5 kJ while Case 2 has an increase of 337.5kJ.
Answer:
The energy needed to provide the increase in speed in the second case is Three times the energy needed to provide increase in speed in the first case.
Explanation:
For case one:
The energy needed for to provide the increase in speed = change in kinetic energy of the car
ΔEk = 1/2mv²- 1/2mu²...................... Equation 1
Where m = mass of the car, v = Final velocity of the car, initial velocity of the car
Given: m = 1000 kg, v = 15 m/s, u = 0 m/s
Substitute into equation 1
ΔEk = 1/2(1000)(15²)- 1/2(1000)(0²)
ΔEk = 112500 J.
For case Two,
Similarly,
ΔEk' = 1/2mv²-1/2mu²...................... Equation 2
Given: m = 1000 kg, v = 30 m/s, u = 15 m/s
ΔEk' = 1/2(1000)(30²)- 1/2(1000)(15²)
ΔEk' = 450000-112500
ΔEk' = 337500 J.
Comparing case one and case two above,
ΔEk' > ΔEk
ΔEk' = 3 ΔEk
Therefore the energy needed to provide the increase in speed in the second case is Three times the energy needed to provide increase in speed in the first case.