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91影视

The string in Fig. 8-38 is L=120cmlong, has a ball attached to one end, and is fixed at its other end. The distancedfrom the fixed end to a fixed peg at point P is 75.0cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?

Short Answer

Expert verified
  1. The speed when it reaches its lowest point 4.85m/s.
  2. The speed when it reaches its highest point is 2.42m/s.

Step by step solution

01

Given data:

Length of the string, L=120cm=1.20m

The distance from the fixed end to a fixed peg at point P, d=75.0cm=0.75m

02

Understanding the concept:

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

03

(a) Calculate the speed when it reaches its lowest point:

As the string reaches its lowest point, its original potential energy is,

U=mgL

(Measured relative to the lowest point) is converted into kinetic energy. Thus,

mgL=12mv2

v=2gL(1)

Here,

The acceleration due to gravity, g=9.80m/s2

Substitute known values in the above equation.

v=2(9.80m/s2)(1.20m)=4.85m/s

Hence, the speed when it reaches its lowest point 4.85m/s.

04

(b) Calculate the speed when it reaches its highest point:

In this case, the total mechanical energy is shared between kinetic 12mvb2and potential mgyb. Note that yb=2r

Where the distance,

r=Ld=(1.200.75)m=0.450m

Energy conservation leads to:

mgL=12mvb2+mgyb12mvb2=mgLmgybvb2=2mg(Lyb)m

vb=2g(L2r)=2(9.80m/s2)(1.20m2(0.450m))=(19.6m/s2)(0.3m)=5.88m2/s2

vb=2.42m/s

Here, the speed when it reaches its highest point is 2.42m/s.

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