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Zorluk: ZorWork, Energy, and Power

A test vehicle accelerating from rest to a speed of 10 m/s10\text{ m/s} requires 50 kJ50\text{ kJ} of net work. Assuming friction and air resistance are negligible, how much additional work must be done on the vehicle to increase its speed from 10 m/s10\text{ m/s} to 20 m/s20\text{ m/s}?

  1. A
    50 kJ50\text{ kJ}
  2. B
    100 kJ100\text{ kJ}
  3. 150 kJ150\text{ kJ}Cevap
  4. D
    200 kJ200\text{ kJ}

Cevap

The additional work required is 150 kJ150\text{ kJ}.
According to the work-energy theorem, net work equals the change in kinetic energy (W=ΔKEW = \Delta KE). Because kinetic energy is proportional to the square of velocity (KE=12mv2KE = \frac{1}{2}mv^2), accelerating from rest to 10 m/s10\text{ m/s} gives the vehicle 50 kJ50\text{ kJ} of kinetic energy. At 20 m/s20\text{ m/s}, the speed is doubled, which quadruples the total kinetic energy to 4×50 kJ=200 kJ4 \times 50\text{ kJ} = 200\text{ kJ}. The additional work required to increase the speed from 10 m/s10\text{ m/s} to 20 m/s20\text{ m/s} is therefore 200 kJ−50 kJ=150 kJ200\text{ kJ} - 50\text{ kJ} = 150\text{ kJ}.

Adım Adım Çözüm

1
Use the work-energy theorem for the initial acceleration from rest to determine the mass-related factor of the vehicle.
W1=ΔKE1=12m(v12−v02)=12m(102−0)=50m=50000 J  ⟹  m=1000 kgW_1 = \Delta KE_1 = \frac{1}{2}m(v_1^2 - v_0^2) = \frac{1}{2}m(10^2 - 0) = 50m = 50{}000\text{ J} \implies m = 1{}000\text{ kg}
The net work done on an object equals its change in kinetic energy.
2
Calculate the total kinetic energy of the vehicle at the final speed of 20 m/s20\text{ m/s}.
KE2=12mv22=12(1000 kg)(20 m/s)2=12(1000)(400)=200000 J=200 kJKE_2 = \frac{1}{2}m v_2^2 = \frac{1}{2}(1{}000\text{ kg})(20\text{ m/s})^2 = \frac{1}{2}(1{}000)(400) = 200{}000\text{ J} = 200\text{ kJ}
Kinetic energy is directly proportional to the square of velocity.
3
Find the additional work by calculating the change in kinetic energy between 10 m/s10\text{ m/s} and 20 m/s20\text{ m/s}.
Wadditional=KE2−KE1=200 kJ−50 kJ=150 kJW_{\text{additional}} = KE_2 - KE_1 = 200\text{ kJ} - 50\text{ kJ} = 150\text{ kJ}
The additional work needed is the difference between final and intermediate energy states.

Anahtar Kavram

Work-Energy Theorem and Velocity-Squared Dependence of Kinetic Energy
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Work, Energy, and Power

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