Question

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.

Answer 1

Answer:

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

$y_i = 1.05 + x$

And final displacement is

$y_f = 0$

Apply the conservation of energy

$\frac{mv_i^2}{2} +\frac{kx_i^2}{2} +mgy_i=\frac{mv_f^2}{2} +\frac{kx^2}{2} +mgy_f$

$\frac{m(0)^2}{2} +\frac{k(0)^2}{2} +mg(1.05+x)=\frac{m(0)^2}{2} +\frac{kx^2}{2} +mg(0)$

$(1.60)(9.81)(1.05 + x)^2=\frac{(315)}{2} x^2\\\\16.4808+15.969x=157.5x^2$

$x^2-0.1014x-0.1046=0$

solve the equation using quadratic equation

$x = \frac{-(-0.1014\pm \sqrt{(-0.1014)^2-4(1)(-1046)} )}{2}$

$x = \frac{0.1014\pm\sqrt{0.42868} }{2} \\\\x = \frac{0.1014\pm0.6547}{2} \\\\x = \frac{0.1014+0.6547}{2} or \frac{0.1014-0.6547}{2} \\\\x = 0.3781 or -0.2767$

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