A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.83 m above the bottom of the chute with an initial speed of 1.91 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 0.269. How far from the bottom of the chute does the toy zebra come to rest

Question

Tyrell

Physics

20 January, 00:22

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.83 m above the bottom of the chute with an initial speed of 1.91 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 0.269. How far from the bottom of the chute does the toy zebra come to rest

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Houston Walsh · Answer 1

The zebra comes to rest at 4.58 m from the bottom of the chute

Explanation:

We are given;

Height; h = 1.83 m

Initial speed; Vi = 1.91 m/s

Coefficient of kinetic friction; = 0.269

Now, from the law of conservation of energy,

Initial Kinetic Energy + Initial Potential Energy = Work done by friction

Now,

Kinetic Energy is given by the formula, KE = (1/2) mv²

Potential Energy is given by the formula; PE = mgh

Also, work done by friction is given as W = µmgd

Thus, we now have;

½mv² + mgh = µmgd

m will cancel out and we have;

½v² + gh = µgd

Plugging in the relevant values to get;

½ (1.91) ² + (9.8 x 1.83) = (0.44 x 9.8) d

1.82405 + 17.934 = 4.312d

d = 19.75805/4.312

d = 4.58 m

Elena Zuniga · Answer 2

Given that,

height h = 1.83 m

Initial speed Vi = 1.91m/s

Coefficient of kinetic friction

uk = 0.269

According to the law of conservation of energy, states that sum of the potential energy and kinetic energy at top of the chute is equal to the kinetic energy at bottom.

∆P. E = ∆K. E

mgh = ½m (Vf²-Vi²)

m cancel out

gh = ½Vf² - ½Vi²

½Vf² = gh + ½Vi²

Vf² = 2gh + Vi²

Vf² = 2*9.81 * 1.83 + 1.91²

Vf² = 39.5527

Vf = √39.5527

Vf = 6.29 m/s

The frictional force exerted on the toy from the bottom of the chute is

Fr = - uk•N

N = W = mg

Fr = ma

Then, ma = - uk•m•g

Divide through by m

a = - uk • g

a = - 0.269 * 9.81

a = - 2.64 m/s²

According to equation of motion to find the distance travel at the bottom

v² = u² + 2as

Note: the final velocity is zero, the body comes to rest. And the initial velocity is the velocity when the zebra reached the ground u = Vf = 6.29m/s

Then, v² = u² + 2as

0² = 6.29² + 2 (-2.64) s

-6.29² = - 5.278s

- cancels out

Then, s = 6.29²/5.27

s = 7.5m

The zebra moves 7.5m from the chute