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1 vote
A sled starts from rest,

and slides down a 38.0°
frictionless hill for 4.24 s.
How fast is it going at
the end?

1 Answer

5 votes

Answer:

25.6 m/s

Step-by-step explanation:

Draw a free body diagram of the sled. There are two forces acting on the sled:

Normal force pushing perpendicular to the hill

Weight force pulling straight down

Take sum of the forces parallel to the hill:

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 38.0°)

a = 6.03 m/s²

Given:

v₀ = 0 m/s

a = 6.03 m/s²

t = 4.24 s

Find: v

v = at + v₀

v = (6.03 m/s²) (4.24 s) + (0 m/s)

v = 25.6 m/s

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