A 68-kg water skier is being pulled by a nylon (Young's modulus 3.7 x 109 N/m2) tow rope that is attached to a boat. The unstretched length of the rope is 15 m and its cross-section area is 1.8 x 10-5 m2. As the skier moves, a resistive force (due to the water) of magnitude 160 N acts on her; this force is directed opposite to her motion. What is the change in length of the rope when the skier has an acceleration whose magnitude is 0.43 m/s2?
To find the change in length of the rope, we need to determine the stress and strain on the rope.
Stress is defined as the force applied on an object per unit area. It can be calculated using the formula:
Stress = Force / Area
In this case, the force acting on the rope is the resistive force of 160 N, and the cross-sectional area of the rope is 1.8 x 10^(-5) m^2. Plugging in these values, we can calculate the stress on the rope:
Stress = 160 N / (1.8 x 10^(-5) m^2) = 8.888 x 10^6 N/m^2
Now, we can use Hooke's Law to relate stress and strain. Hooke's Law states that stress is directly proportional to strain, where strain is the change in length per unit original length of the material.
Hooke's Law: Stress = Young's Modulus x Strain
Rearranging this formula, we get:
Strain = Stress / Young's Modulus
Plugging in the values we have, we can calculate the strain:
Strain = (8.888 x 10^6 N/m^2) / (3.7 x 10^9 N/m^2) = 2.4 x 10^(-3)
Now, we can use the definition of strain to find the change in length of the rope.
Strain = Change in length / Original length
Rearranging the formula, we have:
Change in length = Strain x Original length
The original length of the rope is given as 15 m. Plugging in the values, we can calculate the change in length:
Change in length = (2.4 x 10^(-3)) x 15 m = 0.036 m
Therefore, the change in length of the rope when the skier has an acceleration whose magnitude is 0.43 m/s^2 is 0.036 m.