Question

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 .

Answer 1

Answer:

The answer is "4.97 m".

Explanation:

$u = 1.23\ \frac{m}{s}$

$H= 1.45 \ m$

$\mu = 0.231$

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

$mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}$

$v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)$

   $=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}$

Friction force is given by the formula

$f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g$

$= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}$

Now by using an equation of motion as

$v^2-u^2= 2as$

From the above the distance traveled is

$S=\frac{v^2-u^2}{2a}$

$S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}$

   $=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97$

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