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Energy Conservation Problem W/Ramp

  • Thread starter kmj9k
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
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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
 
  • #3
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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 ?
 
  • #4
radou
Homework Helper
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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: [tex]v = \sqrt{2g(1.5-0.2)}[/tex])
 

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