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 1.07 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 0.285. How far from the bottom of the chute does the toy zebra come to rest? Take g = 9.81 m/s2.
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To find the distance from the bottom of the chute where the toy zebra comes to rest, we need to break the problem into two parts: (1) the motion down the frictionless chute and (2) the motion along the horizontal surface with friction.
1. Motion down the frictionless chute:
Since the chute is frictionless, the only force acting on the zebra is gravity. We can use the laws of kinematics to find the time it takes for the zebra to reach the bottom of the chute. The relevant equation is:
s = ut + (1/2)at²,
where s is the distance traveled, u is the initial speed, t is the time taken, a is the acceleration due to gravity, and using the equation of motion to calculate s.
Given:
Initial height (s) = 1.75 m
Initial speed (u) = 1.07 m/s
Acceleration due to gravity (a) = 9.81 m/s²
Using the equation:
s = ut + (1/2)at²,
we can rearrange to solve for time (t):
1.75 = (1.07)t + (1/2)(9.81)(t²).
Simplifying the equation:
1.75 = 1.07t + 4.905t².
This equation is a quadratic equation. Let's rewrite it in the standard form and solve for t.
4.905t² + 1.07t - 1.75 = 0.
Solving this quadratic equation for t, we get two possible solutions: t = -0.58 s or t = 0.34 s. We discard the negative value because we are interested in the time it takes to reach the bottom.
Therefore, the zebra takes approximately 0.34 seconds to reach the bottom of the chute.
2. Motion along the horizontal surface with friction:
After the zebra emerges from the chute, it continues sliding on a horizontal surface that has friction. The force of friction opposes the motion and will cause the zebra to eventually come to a stop.
To find the distance traveled on the horizontal surface before coming to rest, we can use the equation of motion with constant acceleration:
v² = u² + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
In this case, the initial velocity (u) is the same as the final velocity (since the zebra comes to rest) and is equal to the speed at which the zebra emerges from the chute.
Given:
Initial speed (u) = 1.07 m/s
Coefficient of kinetic friction (μ) = 0.285
The frictional force acting on the zebra is given by:
frictional force = coefficient of friction * normal force.
The normal force is equal to the weight of the zebra, which is given by:
weight = mass * acceleration due to gravity.
First, let's calculate the normal force:
mass = weight / acceleration due to gravity.
mass = (weight of zebra) / 9.81.
Now, let's calculate the frictional force:
frictional force = coefficient of friction * normal force.
Using the frictional force, we can calculate the acceleration (a) acting on the zebra:
frictional force = mass * acceleration.
Next, substitute the values into the equation of motion:
v² = u² + 2as.
Since the zebra comes to rest, the final velocity (v) is zero.
0 = u² + 2as.
Now, solve for the distance traveled (s):
s = -(u²) / (2a).
Substitute the given values into the equation to find the distance traveled:
s = -(1.07²) / [2 * (frictional force / mass)].
Calculate the distance using the given values.
Therefore, the zebra comes to rest approximately [calculate the distance using the equations] meters from the bottom of the chute.