First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂= $\sqrt{43.2)\\$ = 6.57 m/s

5 0

3 years ago

2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?

NeTakaya

Answer:

$\boxed {\tt 3.63636364 \ m/s}$

Explanation:

Velocity can be found using the following formula:

$v=\frac{p}{m}$

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

$p= 200 \ kgm/s\\m=55 \ kg$

Substitute the values into the formula.

$v=\frac{200 \ kg m/s}{55 \ kg}$

Divide. Note that the kilograms, or kg, will cancel each other out.

$v=\frac{200 \ m/s}{55}$

$v= 3.63636364 \ m/s$

The woman's velocity is 3.63636364 meters per second.

6 0

3 years ago