How much work do movers do (horizontally) in pushing a 150-kg crate 12.3 m across a floor at constant speed if the coefficient of friction is 0.70?

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To begin, identify the work done by the movers when moving a crate at constant speed on a horizontal floor: the only opposing force is friction, so the work equals friction force times distance along the displacement.
Option 1 calculation path: The normal force on a horizontal surface is N = mg, since there is no vertical acceleration. With m = 150 kg and g ≈ 9.8 m/s^2, N = 150 × 9.8 = 1470 N. The kinetic friction coefficient is μ = 0.70, so the frictional force is F_f = μN = 0.70 × 1470 = 1029 N (approximately).
Distance is d = 12.3 m, so the work done by the movers against friction is W = F_f × d ≈ 1029 × 12.3 ≈ 12,656.7 J.
Converting to scientific notation, W ≈ 1.27 × 10^4 J, which is very close to the option 1.26 × 10^4 J when rounded.
Option 2: 1845 J is far too small because it would imply a friction force around 150 N over 12.3 m, inconsistent with μ = 0.70 and the given mass.
Option 3: 1.82 × 10^4 J corresponds to a friction force about 1483 N over 12.3 m, which would require μN ≈ 1.21 times larger than calculated, not matching μ = 0.70 and mg.
Option 4: 1300 J suggests a friction force close to 106 J/m, which again contradicts the μ and the weight; the number is simply not compatible with the given parameters.
In summary, the physically correct result given μ = 0.70, m = 150 kg, d = 12.3 m, and g ≈ 9.8 m/s^2 is about 1.26 × 10^4 J, aligning with the first option.

SPHY001 Practice Quiz - Unit 07 | LA047

How much work do movers do (horizontally) in pushing a 150-kg crate 12.3 m across a floor at constant speed if the coefficient of friction is 0.70?

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