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andrey2020 [161]
3 years ago
9

How does work done on an object affect its kinetic energy

Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

According to the work-energy theorem, the work done on an object by a net force equals the change in kinetic energy of the object. Essentially kinetic energy is the energy used for motion. Interestingly, as work is done on an object, potential energy can be stored in that object. A moving object has kinetic energy because work has been done on it. When work is done energy in one form is transferred to the kinetic energy of the moving object. To stop the object again, the same amount of work would have to be done to bring it back to rest.

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Take the measurement of the distance (d) with a meter rule (in meters) and also measure the time (t) of the travel in seconds with a stopwatch.

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The speed of an object in motion is the distance covered by the object with respect to time, that is, the ratio of distance covered to the time taken to reach that distance.

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           = d (in meters m) / t (in seconds s) = m/s

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
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