What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h in a distance of 55 m?
I have tried to work this one and can't seem to get the answer.
2 answers
20kj
The work of the brakes will result in a change in kinetic energy:
W = ΔK
W = ½mv² - ½mv₀²
W = ½m(v² - v₀²)
First, convert units from km/h to m/s:
v₀ = 96km/h (1000m / 1km × 1h / 3600s) = 26.67 m/s
v = 56km/h (1000m / 1km × 1h / 3600s) = 15.56 m/s
Substitute in equation:
W = ½m(v² - v₀²)
W = ½1500kg [(15.56m/s)² - (26.67m/s)²].
W = -351881.475 J ≈ - 3.5 × 10⁵ J
W = ΔK
W = ½mv² - ½mv₀²
W = ½m(v² - v₀²)
First, convert units from km/h to m/s:
v₀ = 96km/h (1000m / 1km × 1h / 3600s) = 26.67 m/s
v = 56km/h (1000m / 1km × 1h / 3600s) = 15.56 m/s
Substitute in equation:
W = ½m(v² - v₀²)
W = ½1500kg [(15.56m/s)² - (26.67m/s)²].
W = -351881.475 J ≈ - 3.5 × 10⁵ J
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