# Conservation of energy - find speed of block

• man_in_motion
In summary: So for homework statement 1 with 0.25m and 5m mass it should read kg not m.In summary, when a 2.4kg block is dropped from a height of 5m, the block experiences a 15cm compression in the spring. The block's speed is 10m/s.
man_in_motion

## Homework Statement

A 2.4kg block is dropped onto a spring from hieght of 5m. When the block is momentarily at rest, the spring is compressed 25cm. Find the speed of the block when the compression of the spring is 15cm.

## Homework Equations

$$K=\frac{1}{2}mv^2$$
$$U_g=mgh$$
$$\Delta U = \Delta K$$

## The Attempt at a Solution

$$U_o=mgh=2.4kg(9.81m/s^2)(5m)=117.72J$$
since all the potential energy gets converted into kinetic energy
$$117.72J=\frac{1}{2}mv^2$$ then I solve for v and plug the numbers in. I get the right answer but I know I'm doing it wrong because I'm not sure how to take the spring into acount, especially when the springs constant isn't given.

Hi man_in_motion!
man_in_motion said:
A 2.4kg block is dropped onto a spring from hieght of 5m. When the block is momentarily at rest, the spring is compressed 25cm. Find the speed of the block when the compression of the spring is 15cm.

$$U_o=mgh=2.4kg(9.81m/s^2)(5m)=117.72J$$
since all the potential energy gets converted into kinetic energy
$$117.72J=\frac{1}{2}mv^2$$ then I solve for v and plug the numbers in. I get the right answer but I know I'm doing it wrong because I'm not sure how to take the spring into acount, especially when the springs constant isn't given.

No, you probably are doing it right.

You don't need to use k, because mgh = 1/2 k 252, and the other PE is 1/2 k 152, so the ks cancel.

thanks tiny-tim, though I'm still not entirley convinced.
consider:
$$K_f-K_o = K_f = U_f-U_o=U_f_g+U_f_s-U_g_o=(mg(0.25m)+\frac{1}{2} \cdot \frac{mg}{0.25m}0.15^2m)-mg(5m)=\frac{1}{2}mv^2$$

Then I solve for v and I get
$$v=\sqrt{\frac{2(mg(0.25)+\frac{1}{2} \cdot \frac{mg}{0.25} \cdot 0.15 - mg(5))}{m}}$$
granted now I no longer get 10m/s, but the answer key could be wrong. what do you think?

Hi man_in_motion!

(just got up :zzz: …)

Sorry, I don't understand your first line.

Energy is conserved
$$K_f+U_f=K_o+U_o$$
$$K_0$$ goes to 0 and rewriting the equation we get $$K_f=U_o-U_f=\Delta U$$
Initially there is only potential energy due to gravity, when the block is in the air
$$U_o_g=mgy=2.4kg(9.81m/s^2)5m$$
There are 2 final potential energies: one for the gravity and one for the spring
$$U_f_g=mgy=2.4kg(9.81m/s^2)(-0.15m)$$
$$U_f_s=\frac{1}{2}kx^2=\frac{1}{2}k(0.15m)^2$$
We don't know k (spring constant) so we need to figure it out using Newton's 2nd law when the block is momentarily at rest

$$\Sigma F_y=ma_y=0=mg-k(0.25)=(2.4kg)(9.81m/s^2)-k(0.25m)=0$$
solving for k I get
$$k=\frac{(2.4kg)(9.81m/s^2)}{(0.25m)}$$

Now we know
$$U_f_s=\frac{1}{2}\frac{mg}{0.25} \cdot (0.15m)^2$$

Plugging things into the first equation

$$K_f=U_f-U_o=mg(0.25m)+\frac{1}{2}\frac{mg}{0.25} \cdot (0.15m)^2-mg(5m) = \frac{1}{2}mv^2$$

sovling for v from the right side of the equation gives
$$v=\sqrt{\frac{2(mg(0.25)+\frac{1}{2}(0.15m)^2-mg(5m))}{m}$$
but this gives me the wrong answer

If I have m following a constant (namley 0.25, 0.15 or 5) it stands for meters (the unit) other wise it's for mass of the block.

Last edited:

## 1. How does conservation of energy apply in finding the speed of a block?

Conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In the case of finding the speed of a block, the initial potential energy of the block is converted to kinetic energy as it moves. By equating these two forms of energy, we can calculate the speed of the block.

## 2. What information is needed to find the speed of a block using conservation of energy?

To calculate the speed of a block using conservation of energy, we need to know the mass of the block, the height from which it was dropped, and the final height at which it stops moving. Additionally, we need to consider any other forces acting on the block, such as friction or air resistance.

## 3. Can conservation of energy be applied in all scenarios to find the speed of a block?

Conservation of energy can be applied in most scenarios to find the speed of a block, as long as the energy is conserved and there are no external forces acting on the block. However, in some cases, such as collisions, we may need to consider other factors such as the coefficient of restitution.

## 4. How does friction affect the calculation of speed using conservation of energy?

Friction is a non-conservative force, meaning it converts some of the initial energy of the block into heat and sound. This means that in order to accurately calculate the speed of the block using conservation of energy, we need to account for the energy lost due to friction.

## 5. What are some real-world applications of using conservation of energy to find the speed of a block?

Conservation of energy is a fundamental principle of physics and is used in a wide range of real-world applications. For example, it can be used to calculate the speed of a roller coaster at different points along its track, or the speed of a projectile launched from a catapult. It is also used in engineering and design to optimize the efficiency of machines and systems.

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