A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.45 m above the bottom of the chute with an initial speed of 1.23 m/s . The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with a coefficient of kinetic friction of 0.231 . How far from the bottom of the chute does the toy zebra come to rest? Assume g=9.81 m/s2 .

Respuesta :

Answer:

The answer is "4.97 m".

Explanation:

[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]

[tex]H= 1.45 \ m\\\\[/tex]

[tex]\mu = 0.231\\\\[/tex]

The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.

[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]

[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]

   [tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]

Friction force is given by the formula

[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]

[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]

Now by using an equation of motion as

[tex]v^2-u^2= 2as[/tex]

From the above the distance traveled is

[tex]S=\frac{v^2-u^2}{2a}[/tex]

[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]

   [tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]

In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]

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