Question

2. A block of mass 1.2 kg lies on a frictionless surface. A man slides the block
against a spring, compressing it .15m. When the man lets go of the spring, the
block moves at 5 m/s. What is the spring constant of the spring?
mm
.

Answer 1

Answer:

The spring constant will be "1333.33 N/m".

Explanation:

The given values are:

Mass,

m = 1.2 kg

Displacement compression,

x = 0.15 m

Block's velocity,

$v_i=5 \ m/s$

As we know,

⇒  $E_i=E_f$

or,

⇒  $K_i+v_i=K_f+v_f$

⇒  $\frac{1}{2}mv_i^2+0=0+ \frac{1}{2}Kx^2$

So,

⇒  $K=\frac{mv_i^2}{x_2}$

On substituting the values, we get

⇒       $=\frac{1.2\times 5\times 5}{0.15\times 0.15}$

⇒       $=\frac{30}{0.0225}$

⇒       $=1333.33 \ N/m$