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Difficulty: EasyWork, Energy and Power

A block is pulled along a smooth horizontal table by a horizontal force of 30 N30\text{ N} through a displacement of 4 m4\text{ m}. At the same time, a vertical upward force of 40 N40\text{ N} acts on the block as it moves. What is the total work done on the block?

  1. 120 J120\text{ J}Answer
  2. B
    280 J280\text{ J}
  3. C
    160 J160\text{ J}
  4. D
    7.5 J7.5\text{ J}

Answer

The total work done on the block is 120 J120\text{ J}.
Work done is given by W=Fdcos⁡θW = F d \cos\theta. The horizontal force acts along the line of motion (θ=0∘\theta = 0^\circ), performing 30 N×4 m=120 J30\text{ N} \times 4\text{ m} = 120\text{ J} of work. The vertical force acts at an angle of 90∘90^\circ to the horizontal displacement, performing zero work because cos⁡(90∘)=0\cos(90^\circ) = 0. Thus, the total work done is 120 J120\text{ J}.

Step-by-Step Solution

1
Identify the formula for work done by a constant force
W=F⋅d⋅cos⁡(θ)W = F \cdot d \cdot \cos(\theta)
Work is defined as the scalar product of force and displacement vectors.
2
Calculate work done by the horizontal force
Whorizontal=30 N×4 m×cos⁡(0∘)=120 JW_{\text{horizontal}} = 30\text{ N} \times 4\text{ m} \times \cos(0^\circ) = 120\text{ J}
The horizontal force is in the exact direction of motion (θ=0∘\theta = 0^\circ).
3
Calculate work done by the vertical force
Wvertical=40 N×4 m×cos⁡(90∘)=0 JW_{\text{vertical}} = 40\text{ N} \times 4\text{ m} \times \cos(90^\circ) = 0\text{ J}
The vertical force is perpendicular to the horizontal displacement (θ=90∘\theta = 90^\circ), so cos⁡(90∘)=0\cos(90^\circ) = 0.
4
Sum the work done by all forces
Wtotal=120 J+0 J=120 JW_{\text{total}} = 120\text{ J} + 0\text{ J} = 120\text{ J}
Work is a scalar quantity, so total work is the algebraic sum of individual work values.

Key Concept

Work done by perpendicular forces is zero
Estimated Time:45s
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Topics

Work, Energy and Power

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