# Conservation of Energy speed of block

• himetx
In summary, a block starting at rest slides down a frictionless track, except for a small rough area, and leaves the track horizontally. It then flies through the air and strikes the ground with a speed of sq rt 2gH m/s. The acceleration of gravity is 9.8 m/s2 and the rough area is 1.2m. The block has a mass of 0.405 kg and initial height of 5.2 m. The final horizontal distance traveled in the air and the total speed at which the block hits the ground can be calculated using the conservation of energy and the work energy theorem. The friction caused by the rough area can be accounted for by finding the work done by friction and subtracting
himetx
A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure attached for#007). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 .

m=0.405 kg
H=5.2 m
h=2 m
rough area=1.2m
μ=0.3

A. What is the speed v of the block when it leaves the track? Answer in units of m/s.

B. What is the horizontal distance x the block travels in the air? Answer in units of m.

C. What is the total speed of the block when it hits the ground? Answer in units of m/s.

The initial height of the block when it is at rest at the top of frictionless track = H

Height of the point where the block leaves the track horizontally = h

Change in height = H - h

Loss in gravitational potential energy of block=mg(H-h)

Gain in kinetic energy of block as it leaves the track =Loss in gravitational potential energy of block

Gain in kinetic energy =(1/2)mv^2

(1/2)mu^2=mg(H - h)

u = sq rt 2g (H - h)

PART A. The ball leaves the track with a speed = sq rt 2g ( H - h)

If H and h are in meter then ball leaves the track with a speed = sq rt19.62 (H-h) m/s

I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

PART C. If V is speed with which block hits the ground.

Final kinetic energy of block (KE) =initial potential energy of block (PE)

(1/2)mV^2=mgH

V= sq rt 2gH

the speed of the block when it hits the ground is sq rt 2gH
If H is in meter then speed is [ sq rt 19.62 H ] m/s

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himetx said:
I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

The conservation law of energy states that energy is never created or destroy, only converted.

When there is friction between two objects, there is a loss in kinetic energy, but this energy doesn't disappear. It's converted into heat (and sound, and sometimes other forms of energy. But only heat here).

$KE_{1} + PE_{1} = KE_{2} + PE_{2} + E_{h}$

Heat energy caused by friction is simply determined by:

$E_{h} = F_{f}d = \mu mgd$

If that doesn't help, let me know.

I don't think heat energy applies to this problem. We have yet to discuss heat energy in my class. If you can show me how this applies using the values given, that would help me understand what your trying to convey.

welcome to pf!

hi himetx! welcome to pf!
himetx said:
I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

you can use conservation of energy, but for the rough patch you need to use the work energy theorem: https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in (mechanical) energy …

so find the work done by https://www.physicsforums.com/library.php?do=view_item&itemid=39", and subtract it from the energy

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As for the horizontal distance traveled by the block in air, it will depend on the angle at which it leaves the track and the initial speed. However, we can use the equation for horizontal displacement, x = ut + (1/2)at^2, to calculate the distance traveled. Since there is no horizontal acceleration, the equation becomes x = ut, where u is the initial horizontal velocity of the block. This can be found using the initial speed calculated in part A and the angle at which the block leaves the track.

## 1. What is the Law of Conservation of Energy?

The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another.

## 2. How does the Law of Conservation of Energy apply to the speed of a block?

According to the Law of Conservation of Energy, the total amount of energy in a system must remain constant. This means that the initial energy of the block, such as potential energy from being lifted, will be converted into kinetic energy as it gains speed.

## 3. Can the speed of a block change without any external force acting on it?

No, according to the Law of Conservation of Energy, the total energy of a system must remain constant. Any change in the speed of a block must be due to an external force, such as friction or a push from another object.

## 4. How does the mass of a block affect its speed?

The mass of a block does not directly affect its speed, but it does affect the amount of force needed to accelerate the block. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. Therefore, a block with a larger mass will require more force to accelerate to a certain speed compared to a block with a smaller mass.

## 5. Can the speed of a block ever exceed the initial energy put into it?

No, the Law of Conservation of Energy states that energy cannot be created. Therefore, the initial energy put into a block, such as potential energy from being lifted, cannot be exceeded. The block may gain additional energy through external forces, but the total energy in the system must remain constant.

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