Problem 41.3 According to Hooke\'s Law, the force (lb), F, required to hold a sp

Question

Problem 41.3 According to Hooke's Law, the force (lb), F, required to hold a spring in place when its displacement from the natural length of the spring is x (ft), is given by the formula F = k x where k is called the spring constant. The value of k varies from spring to spring. Suppose that it requires 120 lb of force to hold a given spring 1.5 ft beyond its natural length. 41.3.1 Find the spring constant for this spring. Include units when substituting the values for F and x into Hooke's Law so that you know the unit on k. 41.3.2 What, including unit, is the constant value of dF/dx 41.3.3 Suppose that the spring is stretched at a constant rate of .032 ft/s. If we define t to be the amount of time (s) that passes since the stretching begins, what, including unit, is the dx constant value of dx/dt? dF 41.3.4 Use the chain rule to find the constant value (including unit) of dF/dt . What is the contextual significance of this value? Problem 41.2 Portions of SW 35th Avenue are extremely hilly. Suppose that you are riding your bike along SW 35th Ave from Vermont Street to Capitol Highway. Let it = d(t) be the distance you have travelled (ft) where t is the number of seconds that have passed since you began your journey. Suppose that e(u) is the elevation (m) of SW 35th Ave where it is the distance (ft) from Vermont St headed towards Capital Highway. 41.2.1 What, including units, would be the meanings of d(25) =300 , e(300) =140 , and (eod)(25)=140 ? 41.2.2 What, including units, would be the meanings of = 41.2.3 Suppose that the values stated in problem 41.1.2 are accurate. What, including unit, is the value of dy/dt | ? What does this value tell you in the context of this problem? =1, value of dt

Explanation / Answer

41.2.1) d(25) = 300 means distance travelled in 25 sec is 300 ft

           e(300)=140 means elevation acheived in when distance travelled is 300ft is 140 m

           eof(25)=140 means e that is elivation is function of d that is distance travelled which is a fuction of time which means this gives the elevation achieved in 25 sec is 140 m

41.2.2) du/dt|t=25 =14 is the rate of change of distance with respect to time or speed i.e. at time 25 sec speed is 14 ft/sec

dy/du|u=300= -0.1rate of change of elevation with respect to distance travelled here elevation decreases with -0.1 m/ft as u increases at the distance 300 ft and it is dimension less

y is a function of u and u is a function of t => using chain rule dy/dt|t=25=(dy/du)*(du/dt)|t=25=(-0.1)*(-14)=1.4 m/s because u = 300ft at t=25 sec and dy/du = -0.1 m/ft at u =300ft

41.3.1) k=F/x = 120lb/1.5ft = 80 lb/ft

41.3.2)dF/dx = d(kx)/dx = k = 80lb/ft

41.3.3)dx/dt= 0.32ft/sec

41.3.4)dF/dt=dF/dx*dx/dt=(80 lb/ft)*(0.32ft/sec)=25.6lb/sec

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