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A boy shoves his stuffed toy zebra, which has mass m, down a frictionless chute, starting at a height D above the bottom of the chute and with an initial speed of v. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction μ. At what distance d from the bottom of the chute does the toy zebra come to rest? Express your answer in terms of the given variables and g, the acceleration due to gravity.

Respuesta :

Manetho

Answer:

[tex]d=\frac{0.5v^2+gD}{\mu\times g}[/tex]

Explanation:

mass of the boy = m kg

staring height of the boy = D meters

initial velocity = v

coefficient of kinetic friction = μ

total energy of the toy is

E_total = K.E. +P.E.

= 0.5mv^2 +mgD

= m( 0.5v^2+ gD)

The work done by friction force

W= μ×F_normal ×distance

W= μ×m×g×d

hence, the distance d covered by the zebra toy

and since, the total energy is conserved

W= E_total

μ×m×g×d= m( 0.5v^2+ gD)

[tex]d=\frac{0.5v^2+gD}{\mu\times g}[/tex]

hence, the distance d from the bottom of the chute does the toy algebra comes to rest.

[tex]d=\frac{0.5v^2+gD}{\mu\times g}[/tex]

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