You drag a crate 2.0 m with a 10 N force at 60° above the horizontal. Set the slider to the work done, using $W = Fd\cos\theta$ ( $\cos 60° = 0.5$ ).

Estimate on a scale from 0 J (0) to 40 J (40).

Hints

Only the force component along the motion does work.
$W = 10 \times 2.0 \times 0.5$ .

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10 (answers within ±1 count)

Why

$W = 10$ J, half the naive $Fd = 20$ J, because half the force aims uselessly upward. At 90° the cosine kills the work entirely: perpendicular forces move no energy.

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You drag a crate 2.0 m with a 10 N force at 60° above the horizontal. Set the slider to the work done, using W=Fdcos⁡θW = Fd\cos\thetaW=Fdcosθ (cos⁡60°=0.5\cos 60° = 0.5cos60°=0.5).

Practise Work, Energy, and the Work-Energy Theorem

More questions on Work, Energy, and the Work-Energy Theorem

You drag a crate 2.0 m with a 10 N force at 60° above the horizontal. Set the slider to the work done, using $W = Fd\cos\theta$ ( $\cos 60° = 0.5$ ).