Question

A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 7.10 cm , then releases the book. What is the book's speed as it slides away? The spring constant is 1550 N/m .

Answer 1

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

spring constant k = 1550 N/m

to find out

book speed

solution

we know energy is conserve so

we can say

loss in spring energy is equal to gain in kinetic energy

so

$\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2$    ..................1

put here value

$\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2$

v = 3.99 m/s

so book speed is 3.99 m/s