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A 60-kilogram sled is coasting with a constant velocity of 10 m/s over smooth ice. It enters a rough stretch of ice 6.0 m longs in which the force of friction is 120 N. a) What is the acceleration during this stretch? B) With what speed dies it emerge from the rough patch?

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MathPhys

Answer:

-2 m/s²

8.7 m/s

Explanation:

Using Newton's second law, the sum of forces in the x direction:

∑F = ma

-120 N = (60 kg) a

a = -2 m/s²

Given:

Δx = 6.0 m

v₀ = 10 m/s

a = -2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (10 m/s)² + 2 (-2 m/s²) (6.0 m)

v = 8.7 m/s

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