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UkoKoshka [18]
3 years ago
10

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

ionless surface. The object is released from rest when the spring is compressed 0.19 m.(a) Find the force on the object.1 N
(b) Find its acceleration at that instant 2 m/s2
Physics
1 answer:
Genrish500 [490]3 years ago
6 0

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

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For more explanation, visit: brainly.com/question/9790340

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Answer:

Time, t = 5.355 seconds

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Distance = 100 m

Initial velocity = 16 m/s

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To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

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Substituting into the equation, we have;

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x_{1} = \frac {10.71}{2}

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Therefore, time = 5.355 seconds

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