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Verdich [7]
3 years ago
8

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi

n, vertical rod. A constant force of magnitude F is being exerted on the rod. The rod remains perpendicular to the direction of the applied force so that the springs are extended by the same amount. Find the effective spring constant k of the two-spring system. Give your answer for the effective spring constant in terms of k1 and k2.
Physics
1 answer:
77julia77 [94]3 years ago
3 0

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

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The average speed of the fielder is 5.24 m/s

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Please, see the attached figure for a graphical description of the problem.

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Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

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With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

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