Find the distance of a frictionless chute when a toy come to rest.

Question

A young girl pushes her toy hippo down a frictionless chute, starting at a height of 1.07 m above the bottom of the chute and with an initial speed of1.81 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 0.219. How far from the bottom of the chute does the toy come to rest? Take 𝑔=9.81 m/s2.

Answer 1

You can also answer this by looking at the total energy energy of the toy. At the start it has Potential Energy PE from the height and also Kinetic Energy KE from the push. At the bottom of the chute all that PE converted to KE and then when it stops the KE is zero. What changed the energy to zero was the work done by friction.

W = ΔK and also W = F D

Put these together and solve for D

FD = ΔK

D = ΔK/F

D = ΔK/umg

note the force of friction = υN =. υmg. since the normal force is equal to the weight (mg).

The initial energy is

Ei = KE + PE = 1/2 mv² + mgh. (the final energy is zero so the change is all the initial energy)

D = (1/2 mv² + mgh) / (u m g). note the mass divides out

D = (1/2 1.81² + 9.81(1.07) )/ ( 0.219 x 9.81)

D = 1.638 + 10.497 / 2.148 = 5.65 m

Answer 2

A young girl pushes her toy hippo down a frictionless chute, starting at a height of 1.07 m above the bottom of the chute and with an initial speed of1.81 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 0.219. How far from the bottom of the chute does the toy come to rest? Take 𝑔=9.81 m/s2.

Vi = 0.81m/s m/s

Vf =?

Distance h = 1.07 m

vf ^2 = Vi^2 +2gh = 0.656 + 20.972 = 21.628

Vf = 4.65 m/s

This is new horizontal velocity

acceleration = - 0.219*9.8

distance=?=D

Again

Vf^2 -Vi^2 = -0.219*9.8 *D

Vf=0

D= 4.65/ (0.219*9.8) = 2.166m