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Elena-2011 [213]
3 years ago
15

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

rack, causes the object to speed up as it goes around. Thework done by the external force as the mass makes one revolution is:
Physics
1 answer:
kodGreya [7K]3 years ago
4 0

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago
A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the
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Answer:

<em>113.4 J</em>

Explanation:

<u>Elastic Potential Energy</u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

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