An archer pulls her bow string back 0.400 m by exerting a force that increases uniformly from zero to 240 N.

(a) What is the equivalent spring constant of the bow?
1 N/m
(b) How much work does the archer do in pulling the bow?
2 J

THANKS!

(a) k = (Maximmum force)/(Maximum deflection)

(b) W = (1/2)(Maximum force)(Maximum deflection)

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To find the answers to these questions, let's go step by step:

(a) To find the equivalent spring constant of the bow, we can use Hooke's Law, which states that the force exerted by a spring (or in this case, the bow) is directly proportional to the displacement from its equilibrium position.

Hooke's Law can be written as: F = k * x

Where F is the force, k is the spring constant, and x is the displacement.

In this case, the force increases uniformly from zero to 240 N, and the displacement is 0.400 m. So we can calculate the spring constant as follows:

240 N = k * 0.400 m

Solving for k:

k = 240 N / 0.400 m

k ≈ 600 N/m

Therefore, the equivalent spring constant of the bow is approximately 600 N/m.

(b) To find the work done by the archer in pulling the bow, we can use the formula for work:

Work (W) = Force (F) * Displacement (d) * cos(theta)

Where F is the applied force, d is the displacement, and theta is the angle between the force and displacement vectors.

In this case, the force is increasing uniformly from zero to 240 N, and the displacement is 0.400 m. Since the angle theta is not given, we can assume it to be zero degrees, as the force and displacement are in the same direction.

So, we can calculate the work done as follows:

W = (240 N) * (0.400 m) * cos(0°)

cos(0°) = 1 (cos of 0 degrees is 1)

W = 240 N * 0.400 m * 1

W = 96 J

Therefore, the archer does 96 J of work in pulling the bow.

Hence, the answers are:
(a) The equivalent spring constant of the bow is 600 N/m.
(b) The archer does 96 J of work in pulling the bow.