Quiz: Work by a Constant Force — 8 questions

Detailed questions and answers

1. A force acts opposite to an object’s displacement. What sign does the work done by that force have?

Undefined
Zero
Negative
Positive

Negative

Explanation

Opposing force and displacement point in opposite directions, which makes their work negative. Positive work occurs when the force and displacement point in the same direction.

2. What quantity is represented by the area under a force-versus-position graph between an initial and a final position?

The force applied at the final position
The displacement between the two positions
The average position during the motion
The work performed by the force

The work performed by the force

Explanation

The area under a force-versus-position graph represents the work performed over the displacement. A force value by itself describes the graph’s vertical value, not the accumulated work.

3. Which statement correctly characterizes the possible sign of work?

Work is a scalar quantity that must have a positive value
Work is a scalar quantity that may be positive or negative
Work is a vector quantity that must point along the displacement
Work is a vector quantity that may point opposite the force

Work is a scalar quantity that may be positive or negative

Explanation

Work is scalar, and its value can be positive or negative depending on the relative directions of force and displacement. Treating work as a vector confuses it with quantities such as force or displacement.

4. An object moves through a distance while a force acts along its motion. Which SI-unit expression represents the resulting work?

m/N\mathrm{m}/\mathrm{N}
Nm\mathrm{N}\,\mathrm{m}
N/m\mathrm{N}/\mathrm{m}
N+m\mathrm{N}+\mathrm{m}

$$\mathrm{N}\,\mathrm{m}$$

Explanation

Work is the product of force and distance, so its units are newtons multiplied by meters, Nm\mathrm{N}\,\mathrm{m}, which is equivalent to a joule. A newton alone represents force rather than work.

5. A constant force has component Fx=6NF_x = 6\,\mathrm{N}, and an object moves from xinitial=2mx_{\mathrm{initial}}=2\,\mathrm{m} to xfinal=7mx_{\mathrm{final}}=7\,\mathrm{m}. What work does the force do?

42J42\,\mathrm{J}
30J-30\,\mathrm{J}
30J30\,\mathrm{J}
9J9\,\mathrm{J}

$$30\,\mathrm{J}$$

Explanation

The displacement is Δx=xfinalxinitial=5m\Delta x=x_{\mathrm{final}}-x_{\mathrm{initial}}=5\,\mathrm{m}, so W=FxΔx=(6N)(5m)=30JW=F_x\Delta x=(6\,\mathrm{N})(5\,\mathrm{m})=30\,\mathrm{J}. Multiplying the force by the final position instead would ignore the initial position.

6. What is the SI unit of work?

The newton, equivalent to one joule-meter
The meter, equivalent to one newton-joule
The watt, equivalent to one newton-meter
The joule, equivalent to one newton-meter

The joule, equivalent to one newton-meter

Explanation

The joule is the SI unit of work and is defined by 1J=1Nm1\,\mathrm{J}=1\,\mathrm{N}\,\mathrm{m}. A newton measures force, while a meter measures distance, so neither one alone is a work unit.

7. A constant force of 12 N12\ \text{N} acts while an object moves 5 m5\ \text{m} in the force’s direction; what does the rectangular area under the force–position graph equal?

5 N5\ \text{N} of force
2.4 J2.4\ \text{J} of work
17 N ⁣ ⁣m17\ \text{N}\!\cdot\!\text{m} of work
60 J60\ \text{J} of work

$$60\ \text{J}$$ of work

Explanation

For a constant force, the rectangular area is F×Δx=12×5=60 JF\times\Delta x=12\times5=60\ \text{J}, which is the work. The rectangle’s height is the force and its width is the displacement, so neither dimension alone gives the work.

8. Which statement best describes mechanical work done by a constant force?

It is the change in force divided by the elapsed displacement
It is the vector sum of force and displacement for the object
It is the ratio of force to displacement along the object’s path
It is the product of force and displacement in the force’s direction

It is the product of force and displacement in the force’s direction

Explanation

Work is calculated from the force and the displacement component in the force’s direction, so it is a scalar result. Force and displacement themselves are vectors, making the vector-sum description incorrect.

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What is work in physics?

The product of a constant force and displacement in the force's direction.

What is the formula for work done by a constant force in x direction?

W=FxΔx=Fx(xfinalxinitial)W = F_x\Delta x = F_x(x_{\mathrm{final}} - x_{\mathrm{initial}})

How is displacement vector expressed for motion along x direction?

Δx=Δxi^\Delta \vec{x} = \Delta x\,\hat{i}

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