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zepelin [54]
3 years ago
13

An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the

object due to its acceleration at left parenthesis 2 Superscript 1 divided by 2 Baseline comma 2 right parenthesis​? ​(Remember Newton's​ law, Fequalsma​.)
Physics
1 answer:
Dmitriy789 [7]3 years ago
7 0

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

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A car weighing 9800 N travels at 30 m/s. What braking force brings it to rest in 100m? In 10 m?
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As per kinematics equation we know that

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initial speed is given as 30 m/s

distance moved = 100 m

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v_f^2 - v_i^2 = 2 a d

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now braking force is given as

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Part b)

Again we have

final speed of the car = 0 m/s

initial speed is given as 30 m/s

distance moved = 10 m

now we have

v_f^2 - v_i^2 = 2 a d

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