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## A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the e

Question

A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

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Physics
1 year
2021-09-01T09:12:11+00:00
2021-09-01T09:12:11+00:00 1 Answers
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## Answers ( )

Answer:a) v = 3.116 m / s, b) μ = 1.65 10⁻²

Explanation:a) to find the velocity of the set, let’s define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

p₀ = M v₀

final instant. After the crash

p_f = (M + m) v

the moment is preserved

M v₀ = (M + m) v

v = [tex]\frac{M}{M+m} \ v_o[/tex]

let’s calculate

v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]

v = 3.116 m / s

b) for this part let’s use the relationship between work and kinetic energy

W = ΔK

as the body has its final kinetic energy is zero

the work of the friction forces is

W = – fr x

the negative sign is because the friction forces always oppose the movement

let’s write Newton’s second law

Y axis

N – W_sled -W_person = 0

N = mg + M g

N = (m + M) g

X axis

fr = ma

the friction force has the expression

fr = μ N

fr = μ g (m + M)

we substitute

– μg (m + M) x = 0- ½ (m + M) v²

μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]

let’s calculate

μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]

μ = 0.0165

μ = 1.65 10⁻²