Question

A 2kg block is dropped from a heights of 40 cm onto a spring constant k=1960N/m.find the maximum distance the spring bis compressed

Answer 1

Answer:

Explanation:

mass of the body is ( m ) = 2 Kg

spring's constant ( K ) = 1960 N/m 

height of the body = 0.4 m

Let the maximum distance spring will compressed = x m

   see the attachment , it is clear that 

initial potential energy of body = mg(h + x) 

final potential energy = spring potential energy = 1/2 Kx² 

                       according to law of conservation of energy,

initial potential energy = final potential energy

 mg(h + x) = 1/2 Kx²

  2mgh + 2mgx = Kx²

 Kx²  - 2mgx - 2mgh = 0

 ⇒ 1960x² - 2*2*9.8*x - 2*2*9.8*0.4 = 0 

⇒ 1960x² - 39.2x - 15.68   = 0

 after solving this quadratic equations , we get x = 0.1m

    hence, answer is x = 0.1m 

Answer 2

m = 2 kg,   h = 0.4 m,   k = 1960 N/m

Let the compression be x. By including it, the potential energy of the block will be,

$\displaystyle\longrightarrow\sf{U=mg(h+x)}$

$\displaystyle\longrightarrow\sf{U=20(0.4+x)}$

This provides the potential energy to the spring due to elasticity, i.e.,

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}kx^2=20(0.4+x)}$

$\displaystyle\longrightarrow\sf{1960x^2=16+40x}$

$\displaystyle\longrightarrow\sf{490x^2-10x-4=0}$

$\displaystyle\longrightarrow\sf{x=\dfrac{10\pm\sqrt{100+7840}}{980}}$

$\displaystyle\longrightarrow\sf{x=1.01\ m\quad\quad;\quad\quad x=-0.08\ m}$

We assumed a positive value for x, and hence,

$\displaystyle\longrightarrow\sf{\underline{\underline{x=1.01\ m}}}$