A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring? .377 Correct: Your answer is correct. m (b) Repeat part (a), but this time assume a constant air-resistance force of 0.750 N acts on the object during its motion.

a) the distance up to which the object compress the spring is x = 0.378m

b),the distance up to which the object compress the spring is x= 0.359

Explanation:

Given that,

Mass , m = 1.6kg

spring at vertical height ,h = 1.05m

Spring constant ,k =  315 N/m

Air resistance force = 0.750N

Let assume that the lowest point is the origin

Therefore, initial displacement is

And final displacement is

Apply the conservation of energy

solve the equation using quadratic equation

Take the positive value

Hence , the distance up to which the object compress the spring is x = 0.378m

b)If the air resistance is taken into cosideration

therefore apply conservation of energy

mg(1.05 + x) - fd = kx²/2

mg(1.05 +x) - (0.750N) (1.05 + x) = kx²/2

((1.60× 9.81) - 0.750)(1.05 + x) = (315/2)x²

14.946 (1.05 + x) = 157.5x²

15.6933 + 14.946x = 157.5x²

x² - 0.0949x - 0.00996

solve the equation using quadratic equation

x = 0.359 or -0.264

Hence ,the distance up to which the object compress the spring is x= 0.359

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