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A boy shoves his stuffed toy zebra down a frictionless chute, starting at a height of 1.75 m above the bottom of the chute and with an initial speed of 1.85 m/s. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction 0.223. How far from the bottom of the chute does the toy zebra come to rest? Take g = 9.81 m/s2.__?___m

Respuesta :

davidlcastiblanco

Answer:

[tex]d=8.63m[/tex]

Explanation:

Given:

[tex]h=1.75m[/tex],[tex]v_i=1.85m/s[/tex],[tex]u=0.223[/tex]

The work of friction is about the kinetic energy and also about the potential energy so:

[tex]W_K=K_E+P_i[/tex]

[tex]u*m*g*d=m*g*h+\frac{1}{2}*m*v^2[/tex]

Cancel the mass as a factor in each element

Solve to d'

[tex]d=\frac{\frac{v^2}{2}+g*h }{u*g}[/tex]

[tex]d=\frac{\frac{1.85^2}{2}*9.8*1.75m}{0.223*9.8}[/tex]

[tex]d=8.63m[/tex]

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