(Figure 1) shows the results of measuring the force F exerted on both ends of a

Question

(Figure 1) shows the results of measuring the force F exerted on both ends of a rubber band to stretch it a distance x from its unstretched position.(Source: www.sciencebuddies.org) The data points are well fit by the equation F=33.55x0.4871, where F is in newtons and x is in meters. (Figure1): https://session.masteringphysics.com/problemAsset/2021960/4/Young14.ch06.p90.jpg

A) Does this rubber band obey Hooke's law over the range of x shown in the graph?

B) The stiffness of a spring that obeys Hooke's law is measured by the value of its force constant k, where k=F/x. This can be written as k=dF/dx to emphasize the quantities that are changing. Define keff=dF/dx and calculate keff as a function of x for this rubber band. For a spring that obeys Hooke's law, keff is constant, independent of x. Express your answer in terms of x.

C) Does the stiffness of this band, as measured by keff, increase or decrease as x is increased, within the range of the data?

D) How much work must be done to stretch the rubber band from x = 0 to x = 0.0400 m? Express your answer to three significant figures and include the appropriate units.

E) How much work must be done to stretch the rubber band from x = 0.0400 to x = 0.0800 m? Express your answer to three significant figures and include the appropriate units.

F) One end of the rubber band is attached to a stationary vertical rod, and the band is stretched horizontally 0.0800 m from its unstretched length. A 0.300-kg object on a horizontal, frictionless surface is attached to the free end of the rubber band and released from rest. What is the speed of the object after it has traveled 0.0400 m? Express your answer with the appropriate units.

Explanation / Answer

keff = dF/dx

now, F = 33.55x^0.4871
dF/dx = 16.3422*x^-0.5129

1. The slope of graph is not constant, so the ubber band does not follow hookes law
2. keff = 16.3422*x^-0.5129
3. The stiffness decreases with increase in x (from the keff eq)
4. dW = Fdx = 33.55x^0.4871 dx
W = 33.55[xf^1.4871 - xi^1.4871]/1.4871 = 0.1936 J
5. W = 33.55[xf^1.4871 - xi^1.4871]/1.4871 = 0.339 J
6. 0.5*0.3*v^2 = 0.339
v = 1.5 m/s

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