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 79.0 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? (c) what maximum force must you apply to move the platform a distance 0.230 m farther?
1. Part A:
The magnitude of the force is calculated using the Hook's law:
You know but you do not have .
You can calculate it using the equation for the work-energy for a spring.
The work done to compress the springs a distance is:
Where is the change in the elastic potential energy of the "spring".
Here you have two springs, but you can work as if they were one spring.
You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:
In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.
Now calculate the magnitude of the force:
2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?
The additional work will be the extra elastic potential energy that the springs earn.
You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.
Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.
3. Part C
What maximum force must you apply to move the platform to the position in Part B?
The maximum force is when the springs are compressed the maximum and that is 0.500m
Therefore, use Hook's law again, but now the compression length is Δx = 0.500m
a) , b) , c)
a) The system can be described by the Principle of Energy Conservation and Work-Energy Theorem:
The work done to compress the springs is:
The force needed to compress the springs is:
b) Spring potential energy is directly proportional to the squared of elongation. Then:
c) The force needed to compress the springs is:
a) F = 680 N, b) W = 215 .4 J , c) F = 1278.4 N
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
= ½ k x²
k = 2 / x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W = -
W = ½ k ( ² - x₀²)
The final distance
= 0.250 +0.220
= 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k
F = 2720 0.470
F = 1278.4 N
a) If you do a work done of about 83.0 J and if we consider that there is no energy lose then the energy stored in the spring will be 83.0 J. So we can write it as,
U = 83.0 J
Energy stored in spring is,
∵ k is the spring constant
∵ x is the distance that tells us how much the spring is compressed or stretched.
We can solve the problem by separately considering the springs or we take these two springs as one spring. So the spring constant ill be
= 2(83)/(0.160)²= 6484 N/m
now the force needed to keep the spring compressed
F = kx = = 1037 N
b) The additional work done can be find out by the change n potential energy due to compression of spring.
Final potential energy =
The additional work 332-83 = 249 J
c) F = k(2x)
= (6484)(2)(0.220) = 2851 N
Energy stored in spring = 1/2 k x²
80 = 1/2 x k .2²
k = 4000 N /m
Required force = k x
= 4000 x .2
Additional work done
= difference in stored elastic energy
= 1/2 k ( x₂² - x₁² )
= .5 x 4000 ( .4² - .2² )
= 240 J
The given data is as follows.
Work done by the force, (W) = 79.0 J
Compression in length (x) = 0.190 m
So, formula for parallel combination of springs equivalent is as follows.
Hence, work done is as follows.
= 4376.73 N/m
Hence, magnitude of force required to hold the platform is as follows.
= 831.58 N
Thus, we can conclude that magnitude of force you must apply to hold the platform in this position is 831.58 N.
F =1490.9 N
work done = 82 J
compression length = 0.220 m
for parallel combination
= 2 k
F = kx
F = 3388.43 ×0.22
F =745.45 N
the additional work done
final potential energy =
= 4 W = 328 J
the additional work = 328 - 82 = 255 J
F = k × 2x
F = 2×745.45
F =1490.9 N