contestada

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest. 1)If the initial speed of the box were doubled, how far x2 would the spring compress?

Respuesta :

Answer:twice of initial value

Explanation:

Given

spring compresses [tex]x_1[/tex] distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1[/tex]

When speed doubles

[tex]\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2[/tex]

divide 1 and 2

[tex]\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}[/tex]

[tex]x_2=2x_1[/tex]

Therefore spring compresses twice the initial value

   

Otras preguntas