A) How fast will it be going when it reaches the lowest point in its swing? B) H

Question

A) How fast will it be going when it reaches the lowest point in its swing?

B) How fast will it be going when it reaches the highest point in its swing (after the string encounters the peg)?

The string in the Figure is L = 118.0 cm long and the distance d to the fixed peg P is 77.9 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. A) How fast will it be going when it reaches the lowest point in its swing? B) How fast will it be going when it reaches the highest point in its swing (after the string encounters the peg)?

Explanation / Answer

A)

Here, at the bottom ,

H = 118 cm = 1.18 m

Using conservation of energy

0.5mv^2 = mgH

0.5 * v^2 = 9.8 * 1.18

v = 4.81 m/s

B)

Now

as r = L -P

then

H = L - 2r

H = L -2(L - P)

H = 2P -L

H = 2*.779 - 1.18

H = 0.378 m

0.5mv^2 = mgH

0.5 * v^2 = 9.8 * 0.378

v = 2.72 m/s

the speed is 2.72 m/s

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