# Energy Conservation Problem W/Ramp

• kmj9k
In summary, the block slides down a frictionless ramp from a height of 1.5 m to 0.20 m, and lands a horizontal distance d away. The velocity at the bottom of the ramp can be found using the equation mgh = (1/2)mv^2, where v = the square root of 2gh. The time it takes for the block to fall can be found using kinematic equations with acceleration being g. The distance traveled can be found using the equation x = initial velocity * square root of (2h)/g. However, an algebraic error may have caused a wrong result. A hint is given to correct the velocity calculation.
kmj9k
Hi, I've been struggling with this problem and would appreciate any help:

A 1.8 kg block slides down a frictionless ramp, as shown in Figure 8-28. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is h = 0.20 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d.

Now, I set the mechanical energy at the top of the ramp equal to the mechanical energy at the bottom of the ramp. Using that equation, I found the velocity of the block at the bottom of the ramp.

From there, I used kinematic equations to find the time it took for the box to fall, with a being g. Then, I used the time to find x. I kept on getting 1.1 m, but the answer is supposed to be 1.02 m. What am I doing wrong?? Thank you in advance!

What is your equation for finding the velocity at the bottom of the ramp?

And what is your equation for find how far it travels?

Your method is right, you probably have an algebraic error

My equation was mgh = (1/2)mv^2, where v = the square root of 2gh. I got 5.4249 m/s. And then for the second part, even if you didn't find the time t, couldn't you also just use the kinematic equations for projectile motion at zero degrees? ie, x= initial V* square root of (2h)/g ?

kmj9k said:
My equation was mgh = (1/2)mv^2, where v = the square root of 2gh. I got 5.4249 m/s. And then for the second part, even if you didn't find the time t, couldn't you also just use the kinematic equations for projectile motion at zero degrees? ie, x= initial V* square root of (2h)/g ?

You got the velocity wrong. (Hint: $$v = \sqrt{2g(1.5-0.2)}$$)

## 1. What is energy conservation?

Energy conservation refers to the practice of reducing the amount of energy used in order to reduce waste and save resources. It involves using energy efficiently and using renewable sources of energy whenever possible.

## 2. What is the "Energy Conservation Problem W/Ramp"?

The "Energy Conservation Problem W/Ramp" refers to a common physics problem that involves calculating the potential energy and kinetic energy of an object on a ramp. It is often used to demonstrate the principles of energy conservation.

## 3. How does energy conservation relate to the ramp problem?

In the ramp problem, the energy of the object is conserved as it moves from a higher position to a lower position, in accordance with the law of conservation of energy. This demonstrates the importance of conserving energy in our daily lives.

## 4. What are some real-life examples of energy conservation on a ramp?

Some real-life examples of energy conservation on a ramp include using a bicycle to go up and down hills, using a wheelchair ramp to conserve energy for individuals with limited mobility, and using ramps in loading docks to conserve energy for workers loading and unloading heavy objects.

## 5. How can we apply the concept of energy conservation on a ramp in our daily lives?

We can apply the concept of energy conservation on a ramp in our daily lives by using ramps or slopes to move objects instead of using energy-intensive methods like lifting or carrying. We can also make sure to turn off lights and appliances when not in use, use public transportation or bike instead of driving, and choose energy-efficient products and appliances.

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