A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 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 0.255 . How far from the bottom of the chute does the toy zebra come to rest
2 answers:
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
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Answer:
Same direction: t=234s; d=6.175Km
Opposite direction: t=27.53s; d=0.73Km
Explanation:
If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be (<em>the train must see the car advancing at a lower speed</em>), where
is the speed of the automobile and
the speed of the train.
So we have .
So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:
And in that time the car would have traveled (<em>relative to the ground</em>):
If they are traveling in opposite directions, <u>we have to do all the same</u> but using (<em>the train must see the car advancing at a faster speed</em>), so repeating the process: