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Komok [63]
3 years ago
8

A boy, standing at point O, is whirling a ball on a string over his head. After several whirls, he lets go of the string and the

ball travels in a straight line. Describe the velocity of the ball in both situations: whirling overhead and being let go.
Physics
2 answers:
emmasim [6.3K]3 years ago
6 0

It will continue in a straight line.

Vikki [24]3 years ago
4 0
When a ball is whirled using a string, it is restricted to move only in circular motion because the net force acting on the ball is towards the center of the circle. Hence, the acceleration of the ball is towards the center. But the velocity of ball is tangential to this circular path all the time. When the whirling is stropped, the string becomes slack and tension in the string becomes zero. The ball no more performs circular motion and the ball moves tangentially to the circle in straight line. Therefore, before letting go, velocity was variable. After letting go, velocity becomes constant.
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8 0
3 years ago
Read 2 more answers
2
Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

\displaystyle U_i = K_f

Substitute and solve for final velocity:
\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\  2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ &  =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

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Answer:

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Plus I took the test and got it right

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Answer:

Explanation:

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