Find elastic energy in the compressed spring.

[paulimerci]

Homework Statement

A massless spring with force constant k = 400N/m is fastened at its left end to a vertical wall, as shown in figure 1. Initially, block C(mass m_c =4.0kgs) and block D )mass m_D=2.0kgs) rest on a rough horizontal surface with block C in contact with the spring (but not compressing it) and with block D in contact with block C. Block Cis then moved to the left, compressing the spring of a distance of 0.50m, and held in place while block D remains at rest.
a) Determine the elastic energy stored in the compressed spring.
Block C is then released and accelerates to the right, toward block D. The surface is rough and the coefficient of friction between each block and the surface is 0.4. The two blocks collide instantaneously, stick together, and move to the right at 3m/s. Remember that the spring is not attached to block C. Determine each of the following.
b) the speed v_c of block C just before it collides with block D
c) the horizontal distance the combined blocks move after leaving the spring before coming to rest.

Relevant Equations

U_s = 1/2 kx^2
Conservation of energy

a) Elastic potential energy stored in the compressed spring is written by, where k =400N/m, compressed spring distance x = 0.5m
$$ U_g = \frac {1}{2}kx^2$$
$$ U_g = 50J$$
b) When block C is compressed, it has stored spring PE and when it is released, the block accelerates to the right, where it is said the surface is rough and some energy is lost in the form of heat. And so the stored spring PE is transformed to K.E + W_NC. I'm using conservation of energy to solve this part. Initial energies when the block C is compressed are U_g and the final energies when the block is released are K.E_f and loss of energy due to heat.
$$ E_i = E_f$$
$$ U_g = K.E_f + W_{NC}$$
$$ \frac{1}{2}kx^2 = \frac {1}{2}m_1v_c^2 + F_f x$$
$$\frac {1}{2}kx^2 - F_f x = \frac{1}{2}m_1v_c^2$$
$$ v_c = 4.59 m/s$$
c) As soon as the block C is released, it accelerates toward the block D the two blocks collide instantly and stick together, which implies that the kinetic energy before the collision is not equal to the kinetic energy after it, and so it is an inelastic collision. I'm using conservation of energy, where the initial energies are when the block C is released, it possesses K.E initial, and the final energies are K.E final and W_NC when the block C collides with block D and that the spring is not attached to block C, initial and final energies will not have $U_g$ where $v_f = 3m/s$, $m_1 = 4kg$, $m_2 = 2kg$,
$$ E_i = E_f$$
$$ K.E_i = K.E_f + W_NC$$
$$\frac{1}{2}m_1 v_c^2 = \frac {1}{2}(m_1+m_2)v_f^2 + F_f d$$
$$\frac{1}{2}m_1 v_c^2 = \frac{1}{2} (m_1+m_2)v_f^2 + \mu g(m_1+m_2) d$$
$$ d = 0.48m$$

Have I done it right?

 

[erobz]

Part c has some issues. The problem dictates the blocks have speed 3 m/s instantaneously after collision. You are supposed to find how far they have traveled after all that kinetic energy has been converted to heat.

 

[paulimerci]

Part c has some issues. The problem dictates the blocks have speed 3 m/s instantaneously after collision. You are supposed to find how far they have traveled after all that kinetic energy has been converted to heat.

Okay, any hints?

 

[erobz]

Okay, any hints?

What is the kinetic energy of the blocks immediately after the collision? The force of friction does that much work in stopping the blocks.

 

[paulimerci]

The kinetic energies of the blocks after collision is #\frac{1}{2}(m_1+m_2)v_f^2##. That’s what I’ve written in post #1.

 

[erobz]

The kinetic energies of the blocks after collision is #\frac{1}{2}(m_1+m_2)v_f^2##. That’s what I’ve written in post #1.

Correct, but you have the initial KE as $\frac{1}{2}m_1v_c^2$? You are kind of (incorrectly) working before and after the collision, when you should be working entirely after the collision has occurred until both masses have no kinetic energy.

 

[paulimerci]

Correct, but you have the initial KE as $\frac{1}{2}m_1v_c^2$? You are kind of (incorrectly) working before and after the collision, when you should be working entirely after the collision has occurred until both masses have no kinetic energy.

I assumed that the velocity of block C to be v c because I reasoned that before block C and block D contact, block C will have some kinetic energy.

 

[erobz]

I assumed that the velocity of block C to be v c because I reasoned that before block C and block D contact, block C will have some kinetic energy.

I think you are missing the point. block C is moving with the block D at 3 m/s initially. You are getting mixed up with energies before and after the collision, when you need to only consider what is happening after the collision. They are indirectly telling you how much energy was lost instantaneously in the collision, by telling you the stick and move together at 3 m/s, but it is of no consequence to the question posed in part c.

 

[haruspex]

c) As soon as the block C is released, it accelerates toward the block D the two blocks collide instantly and stick together, which implies that the kinetic energy before the collision is not equal to the kinetic energy after it, and so it is an inelastic collision. I'm using conservation of energy, where the initial energies are when the block C is released, it possesses K.E initial, and the final energies are K.E final and W_NC when the block C collides with block D and that the spring is not attached to block C, initial and final energies will not have $U_g$ where $v_f = 3m/s$, $m_1 = 4kg$, $m_2 = 2kg$,
$$ E_i = E_f$$
$$ K.E_i = K.E_f + W_NC$$
$$\frac{1}{2}m_1 v_c^2 = \frac {1}{2}(m_1+m_2)v_f^2 + F_f d$$
$$\frac{1}{2}m_1 v_c^2 = \frac{1}{2} (m_1+m_2)v_f^2 + \mu g(m_1+m_2) d$$
$$ d = 0.48m$$

For part c, you need to be more methodical.
What is your initial state and what is your final state?
Write the total mechanical energy expression for each of those states.
Then write the equation relating the initial state mechanical energy, the final state mechanical energy and the work lost to friction.

 

[paulimerci]

I think you are missing the point. block C is moving with the block D at 3 m/s initially. You are getting mixed up with energies before and after the collision, when you need to only consider what is happening after the collision. They are indirectly telling you how much energy was lost instantaneously in the collision, by telling you the stick and move together at 3 m/s, but it is of no consequence to the question posed in part c.

Okay I understood what mistake I'm doing,
$$ K.E_i = K.E_f + W_NC$$
$$\frac{1}{2} (m_1+m_2)v_f^2 = 0+ \mu g(m_1+m_2) d$$
$$ d= 1.15m$$

 

[paulimerci]

For part c, you need to be more methodical.
What is your initial state and what is your final state?
Write the total mechanical energy expression for each of those states.
Then write the equation relating the initial state mechanical energy, the final state mechanical energy and the work lost to friction.

Yes, thank you!

 

[paulimerci]

Okay I understood what mistake I'm doing,
$$ K.E_i = K.E_f + W_NC$$
$$\frac{1}{2} (m_1+m_2)v_f^2 = 0+ \mu g(m_1+m_2) d$$
$$ d= 1.15m$$

Thank you @erboz @haruspex!

 

[erobz]

Hey, for future reference for latex: double or larger subscripts go inside braces. I see you did this a few times $W_NC$ when it should be $W_{NC}$.