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Zorluk: KolayWork, Energy and Power

A constant horizontal force of 25 N25\text{ N} is applied to push a box across a smooth horizontal surface through a displacement of 8 m8\text{ m} in the direction of the force. What is the total work done on the box?

  1. 200 J200\text{ J}Cevap
  2. B
    40 J40\text{ J}
  3. C
    3.125 J3.125\text{ J}
  4. D
    33 J33\text{ J}

Cevap

The work done on the box is 200 J200\text{ J}.
Work done is defined as the product of force and distance moved in the direction of the force (W=F×dW = F \times d). Substituting the given values F=25 NF = 25\text{ N} and d=8 md = 8\text{ m} gives W=200 JW = 200\text{ J}.

Adım Adım Çözüm

1
Identify the given physical quantities
Force F=25 NF = 25\text{ N}, displacement d=8 md = 8\text{ m}, angle θ=0∘\theta = 0^\circ
Work depends on force and displacement along the line of action
2
Apply the work formula W=F⋅dcos⁡(θ)W = F \cdot d \cos(\theta)
W=25 N×8 m×cos⁡(0∘)=200 JW = 25\text{ N} \times 8\text{ m} \times \cos(0^\circ) = 200\text{ J}
Since force and displacement are in the same direction, cos⁡(0∘)=1\cos(0^\circ) = 1

Anahtar Kavram

Work Done by a Constant Force
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Work, Energy and Power

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