Examkin

Soru

Zorluk: OrtaWork, Energy, and Power

A pneumatic test probe is launched vertically upward from ground level with an initial kinetic energy of 450 J450\text{ J}. Assuming air resistance is negligible, what is the kinetic energy of the probe at the point where it has gained 300 J300\text{ J} of gravitational potential energy?

  1. A
    300 J300\text{ J}
  2. 150 J150\text{ J}Cevap
  3. C
    750 J750\text{ J}
  4. D
    0 J0\text{ J}

Cevap

150 J150\text{ J}
In the absence of air resistance, the total mechanical energy of the system remains constant (E=KE+PE=450 JE = KE + PE = 450\text{ J}). When the probe gains 300 J300\text{ J} of gravitational potential energy, the remaining kinetic energy is 450 J−300 J=150 J450\text{ J} - 300\text{ J} = 150\text{ J}.

Adım Adım Çözüm

1
Determine the total initial mechanical energy of the probe at launch.
Total Mechanical Energy (EE) = 450 J450\text{ J}
At ground level before rising, all mechanical energy is kinetic energy (KE=450 JKE = 450\text{ J}, PE=0 JPE = 0\text{ J}).
2
Apply the law of conservation of mechanical energy.
E=KE+PE=450 JE = KE + PE = 450\text{ J}
In the absence of air resistance, the total mechanical energy of the system remains constant throughout the upward climb.
3
Calculate the remaining kinetic energy at the given potential energy level.
KE=450 J−300 J=150 JKE = 450\text{ J} - 300\text{ J} = 150\text{ J}
Subtract the gained gravitational potential energy from the total conserved mechanical energy.

Anahtar Kavram

Conservation of Mechanical Energy
Tahmini Süre:1m 0s
Bu soruyu puanla

Konular

Work, Energy, and Power

Bu konudan daha fazla soru