 Matter & Interactions 2nd ed. Practice Problems Aaron Titus | High Point University home
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 1570001     Work done on a springs in a piece of exercise equipment     1570001 View Question | View Solution | Download pdf Question The piece of exercise equipment shown below includes two handles that are on each end of two parallel springs, each with a stiffness 500 N/m. To build strength, a person holds the two handles and pulls them apart with equal magnitude forces applied in opposite directions to the handles, thus stretching the springs. Neglect the mass of the springs and assume that they are pulled horizontally. Figure: Two parallel springs between two handles that can be pulled apart. How much work must a person do to stretch the springs 5 cm? Suppose that a person is already holding the handles so that the springs are stretched 5 cm from their relaxed lengths. How much more work must the person do to stretch the springs an additional 5 cm (i.e. from 5 cm stretched to 10 cm stretched)?

 Solution Define the system to be the two springs. The important point to realize is that if the springs stretch a total of 5 cm, then the person's left hand may have moved 2.5 cm and the person's right hand may have moved 2.5 cm. This is the same as if one hand stayed still, and the other hand moved 5 cm. So to simplify things, assume that one hand stays at rest, and the other hand moves 5 cm, just like one end of the spring is connected to a wall. Assume that the left handle remains at rest. Define the coordinate system so that x=0 is the relaxed (unstretched) position of the right handle. In this way, the distance stretched, s is the position of the right handle, x. Figure: The coordinate system. The work done on one spring by the right hand is Since there are two springs of the same stiffness that stretch the same amount, the right hand also does work on the other spring. Thus, the total work done by the right hand is (b) If the spring is already stretched 5 cm and is then stretched 5 more cm, then the work done on one spring during this interval is The total work done on both springs is twice the work done on one spring Note that the work done on the spring to stretch it from 0.05 m to 0.10 m is greater than the work done to stretch the spring from 0 to 0.05 m. Though the displacement of the handle is the same in each case (5 cm), the average the force on the spring is greater when the spring is stretched from 0.05 m to 0.10 m. Thus more work is required to stretch it through this interval.  