Question

A 2.7-kg block slides with a speed of 1.1 m/s on a frictionless, horizontal surface until it encounters a spring.

Answer 1

Answer

given,

mass of block = 2.7 kg

speed of block = 1.1 m/s

a) spring compressed to distance of 6 cm or 0.06 m

  using conservation of energy

kinetic energy is converted into spring energy

$\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2$

$k =\dfrac{ mv^2}{x^2}$

$k =\dfrac{2.7\times 1.1^2}{0.06^2}$

k= 907.5 N/m

b) again using conservation of energy

kinetic energy is converted into spring energy

$\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2$

$v =\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{907.5\times 0.015^2}{2.7}}$

$v=\sqrt{0.075625}$

v= 0.275 m/s