Question

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is dropped onto the spring.
(a) How far does the object compress the spring?
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.
m
(c) How far does the object compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s2 and air resistance is neglected?
m

Answer 1

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

Answer 2

The compression of the spring when it is dropped from 1.05 m is 0.37 m.

The compression of the spring when air resistance is considered 0.36 m.

The compression of the spring when air resistance is neglected and gravity is 1.63 is 0.14 m.

The given parameters;

• mass of the object, m = 1.6 kg
• initial position of the object, h₀ = 1.05 m
• spring constant, k = 330 N/m

The compression of the spring is determined by applying the principle of conservation of energy;

$\frac{1}{2} kx^2 = mgh\\\\\frac{1}{2} kx^2 = mg(1.05 + x)\\\\kx^2 = 2mg(1.05 + x)\\\\330x^2 = 2\times 1.6 \times 9.8(1.05 + x)\\\\330x^2 = 32.93 + 31.36x\\\\330x^2 - 31.36x - 32.93 = 0\\\\a = 330, \ b = -31.36, \ c = -32.93\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\x = \frac{-(-31.36) \ \ +/- \ \sqrt{(-31.36)^2 -4(330\times -32.93)} }{2(330)}\\\\x = 0.37 \ m$

Considering air resistance, the compression of the spring is calculated as follows;

$\frac{1}{2} kx^2 = mg(1.05+ x) - F(1.05+ x)\\\\\frac{1}{2} \times 330 x^2 = 1.6\times 9.8(1.05 + x) - 0.75(1.05 + x)\\\\165x^2 = 16.46 + 15.68x - 0.79 - 0.75x \\\\165x^2 -14.93x - 15.67 = 0\\\\a = 165, \ \ b = -14.93 \ \ c = -15.67 = 0\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-14.93) \ \ +/- \ \sqrt{(-14.93)^2 - 4(15.67)} }{2(165)}\\\\x = 0.36 \ m$

The compression of the spring when air resistance is neglected and gravity is 1.63;

$\frac{1}{2} kx^2 = mg(1.05 + x)\\\\\frac{1}{2} \times 330 x^2 = 1.6 \times 1.63(1.05 + x)\\\\165 x^2 = 2.74 + 2.61x \\\\165 x^2 - 2.61x- 2.74 = 0\\\\a = 165, \ \ b = -2.61, \ c = \ -2.74\\\\x = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-2.61) \ \ +/- \ \ \sqrt{(-2.61) ^2 - 4(165\times -2.74)} }{2(165)}\\\\x = 0.14 \ m$

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