Bullets

[hauthuong]

A bullet of mass 0.0021 kg is shot into a wooden block of mass 0.197 kg.

They rise to a final height of 0.546 m as shown. What was the initial speed (in m/s) of the bullet before it hit the block?

I figured out this one
I got another one
A 5.0 kg ball is attached to a vertical unstretched spring with k = 762 N/m and then released. What distance does the ball fall just as it momentarily comes to rest?
I use F=-kx with F=mg therefore mg=-kx from there I solve for x but I got the wrong answer. What am I wrong. Thank you

[Andrew Mason]

A 5.0 kg ball is attached to a vertical unstretched spring with k = 762 N/m and then released. What distance does the ball fall just as it momentarily comes to rest?
I use F=-kx with F=mg therefore mg=-kx from there I solve for x but I got the wrong answer. What am I wrong. Thank you
Think of it as a conversion of gravitational potential energy into spring potential energy and ball kinetic energy, as the ball falls.
mgx = \frac{1}{2}(kx^2 + mv^2)

Since maximum displacement occurs when v=0,
mgx = \frac{1}{2}kx^2

x = 2mg/k

AM

[hauthuong]

thank you
A 1.72 kg object is suspended from a spring with k = 19.1 N/m. The mass is pulled 0.305 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinentic energy of the object in J?
1/2 kx^2=1/2 mv^2
and I got it wrong too

[Yegor]

Try to use 1/2 kx^2=1/2 mv^2+mgl, l - heigt
Because when the mass oscillates, it's potential energy changes

[Andrew Mason]

thank you
A 1.72 kg object is suspended from a spring with k = 19.1 N/m. The mass is pulled 0.305 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinentic energy of the object in J?
1/2 kx^2=1/2 mv^2
and I got it wrong too
The condition for maximum kinetic energy is dK/dt = 0 (when the rate of change of kinetic energy = 0). Since K = \frac{1}{2}mv^2 = \frac{1}{2}m(dx/dt)^2 the maximum kinetic energy occurs when d^2x/dt^2 = 0 (ie. when a = 0).

The equation of motion is:
F = mg-Kx = ma where x = the displacement from equilibrium

So when a=0
kx=mg
x=mg/k

Note that it is independent of the maximum amplitude. To find the speed when x=mg/k, use an energy approach:
U_g + KE + U_k = U_{ki}
mg(A-x) + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2


AM

[Yegor]

I'm afraid, that I didn't understand all correctly, but the question is "What is the maximum kinentic energy of the object in J?", isn't it?
I think You are right speaking about condition for maximum kinetic energy, however it depends on amplitude x.

1/2 kx^2=1/2 mv^2+mgx (conversion of energy)

1/2 mv^2=x(kx/2-mg)

[Andrew Mason]

I'm afraid, that I didn't understand all correctly, but the question is "What is the maximum kinentic energy of the object in J?", isn't it?
I think You are right speaking about condition for maximum kinetic energy, however it depends on amplitude x.

1/2 kx^2=1/2 mv^2+mgx (conversion of energy)

1/2 mv^2=x(kx/2-mg)
see my edited reply above.

AM

[Yegor]

But i have
mgx+(1/2)mv^2=(1/2)k(A-x)^2,
where A=(mg/k+x)
What do You think?

[Andrew Mason]

But i have
mgx+(1/2)mv^2=(1/2)k(A-x)^2,
where A=(mg/k+x)
What do You think?
A= initial amplitude.
Try:
mg(A-x) + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2

substituting x=mg/k:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2 - mg(A - mg/k) - \frac{1}{2}m^2g^2/k

v^2 = KA^2/m - 2gA + mg^2/k
v = \sqrt{KA^2/m - 2gA + mg^2/k}

AM