Question

ou place the spring vertically with one end on the floor. You then drop a book of mass 1.40 kgkg onto it from a height of 0.800 mm above the top of the spring. Find the maximum distance the spring will be compressed.

Answer 1

Answer:

The maximum distance compressed by the spring = 0.126 m

Explanation:

k = 1600 N/m

Given,

Mass of book = 1.40 kg

Height book above the spring = 0.80 m

Let the maximum distance the spring will be compressed be x

The change in potential energy of the book = workdone in compressing the spring by a Maximum distance of x

Change in potential energy of the book moving through height 0.80 m and a distance of x is given as

= mg(0.8 + x) = 1.40 × 9.8(0.8+x)

= 13.72(0.8+x) = (10.976 + 13.72x) J

Workdone in compressing the spring by a distance of x = (1/2)(1600)(x²) = (800x²) J

10.976 + 13.72x = 800x²

800x² - 13.72x - 10.976 = 0

Solving the quadratic equation

x = 0.126 m or - 0.108 m

The answer that supports the question according to the laws of Physics = 0.126 m

The maximum distance compressed by the spring = 0.126 m

Hope this Helps!!

Answer 2

Complete question:

A spring of negligible mass has force constant k = 1600 N/m. (a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.40-kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

Answer:

(a) 0.063 m

(b) 0.126 m

Explanation:

Given;

force constant, K =  1600 N/m

Part (a)

Elastic potential energy is given as;

U = ¹/₂Kx²

where;

x is the extension in the spring

$x = \sqrt{\frac{2U}{K} } = \sqrt{\frac{2*3.2}{1600} } = 0.063 \ m$

Part (b)

given;

mass of the book, m = 1.4 kg

height above the spring from which the book was dropped, h = 0.8 m

From the principle of conservation of energy;

Gravitational potential energy = Elastic potential energy

mgH = ¹/₂Kx²

H is the total vertical distance from floor to 0.8 m =  maximum distance the spring will be compressed + h

let the maximum distance = A

mg(A+h) = ¹/₂KA²

1.4 x 9.8(A + 0.8) = ¹/₂ x 1600A²

13.72 (A + 0.8) = 800A²

13.72A + 10.976 = 800A²

800A² - 13.72A -  10.976 = 0

This is a quadratic equation, and we solve using formula method, where a = 800, b = - 13.72 and c = - 10.976

A = 0.126 m