Question

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

Answer 1

Answer:

$d=8.63m$

Explanation:

Given:

$h=1.75m$,$v_i=1.85m/s$,$u=0.223$

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

$W_K=K_E+P_i$

$u*m*g*d=m*g*h+\frac{1}{2}*m*v^2$

Cancel the mass as a factor in each element

Solve to d'

$d=\frac{\frac{v^2}{2}+g*h }{u*g}$

$d=\frac{\frac{1.85^2}{2}*9.8*1.75m}{0.223*9.8}$

$d=8.63m$