A force of 34 pounds is required to hold a spring stretched 2.3 feet beyond its

Question

A force of 34 pounds is required to hold a spring stretched 2.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 4.2 feet beyond its natural length? It took 34 Joules of work to stretch a spring from its natural length of 14 meters beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance 16 meters? Find the x-coordinate x_c, or the y-coordinate y_c of the centroid of the region bounded by the given curves. y = -x + 2, 0 lessthanorequalto x lessthanorequalto 2 y = e^2x, y = 0, x = 0, x = 2

Explanation / Answer

1)

1)F=kx

k=F/x

k=34/2.3

work done = [0 to 4.2] (34/2.3)x dx

work done = [0 to 4.2] (34/2.3)(1/2)x2

work done = (34/2.3)(1/2)4.22 -0

work done = 130.4 pond feet

2)

work=(1/2)kx2

34=(1/2)k 142

k=34*2/142

k=17/49

x =16 m

F=kx

F=(17/49)*16

F=5.55 Newton

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