Question

A bead slides on the frictionless wire as shown in the figure attached. If the speed of the beead is 2.0 m/s when its at A, how fast will the bead be going at point B and at point C?

Answer 1

Answer on Question #62081-Physics-Mechanics-Relativity

A bead slides on the frictionless wire as shown in the figure attached. If the speed of the bead is $2.0 \, \text{m/s}$ when it is at A, how fast will the bead be going at point B and at point C?

Solution

$$\frac {m v ^ {2}}{2} + m g h = c o n s t$$ $$\frac {v ^ {2}}{2} + g h = c o n s t$$

At B:

$$\frac {v ^ {2}}{2} = \frac {2 ^ {2}}{2} + (9. 8) (1)$$ $$v = 4. 9 \frac {m}{s}.$$

At C:

$$\frac {v ^ {2}}{2} = \frac {4 . 9 ^ {2}}{2} - (9. 8) (0. 8)$$ $$v = 2. 9 \frac {m}{s}.$$

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