Leg Presses. As part of your daily workout, you lie on your back and push with y
ID: 1779633 • Letter: L
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Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 81.0 J when you compress the springs a distance of 0.230 m from their uncompressed length.
A) What magnitude of force must you apply to hold the platform in this position?
B) How much additional work must you do to move the platform a distance 0.230 m farther?
B) How much additional work must you do to move the platform a distance 0.230 m farther?
C) What maximum force must you apply to move the platform to the position in Part B? Problem 6.34 Part A Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 81.0 J when you compress the springs a distance of 0.230 m from their uncompressed length What magnitude of force must you apply to hold the platform in this position? Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining; no points deducted Part B How much additional work must you do to move the platform a distance 0.230 m farther? Submit My Answers Give Up Part C What maximum force must you apply to move the platform to the position in Part B? Submit My Answers Give UpExplanation / Answer
(a) If you do 81.0 J of work then, assuming no energy loss (due to friction for example), there will be 81.0 J of energy stored in the springs.
The potential energy, U = 81.0 J, stored in a spring is equal to
U = 1/2 kx^2 where k is the spring constant and x is distance the spring is stretched or compressed. The problem could be done treating the springs separately and assuming they are the same, or you can treat the two springs as one spring with twice the spring constant as each individual spring. Treating the springs as one spring the effective spring constant is
k = 2U / x^2 = (2 * 81)/0.230^2 = 3062.38 N/m
The force needed to keep the springs compressed by 0.230m is
F = kx = 3062.38*0.230 = 704.35 N
(b) The additional work can be calculated by determined the change in the potential energy stored in the spring by the additional compression.
The final potential energy = (½)k(2x)^2 = 4U= 4*81 = 324 J
The additional work is then 324 - 81 = 243 J
(c) The maximum force you have to apply to compress it an additional 0.230m is the force required to keep the springs compressed by 0.460m
F = k * (2x) = (3062.38)*(0.460) = 1408.69 N
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