Calculating Final Speed of a Crate on a Ramp with Initial Velocity of 0.6 m/s

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The discussion focuses on calculating the final speed of a crate sliding down a ramp with an initial velocity of 0.6 m/s. The crate, weighing 4.00 kg, slides down a 1.80 m ramp inclined at 28.0° and experiences a friction force of 4.50 N. Initially, the user calculated the speed at the bottom of the ramp to be 3.54 m/s and the distance traveled on the horizontal floor to be 5.57 m. When faced with the new scenario of an initial speed, the user was advised to include the initial kinetic energy in the energy equation, which resolved their confusion. This approach successfully led to the correct calculation of the final speed.
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"A 4.00-kg crate slides down a ramp. The ramp is 1.80 m in length and inclined at an angle of 28.0° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 4.50 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
https://www.webassign.net/serpse8/8-figure-10-alt.gif (figure)
(A) Use energy methods to determine the speed of the crate at the bottom of the ramp.

(B) How far does the crate slide on the horizontal floor if it continues to experience a friction force of magnitude 4.50 N?"so i solved part A for the speed of the crate at the bottom of the ramp and got 3.54m/s.

Using the speed, i then solved part B for the distance it traveled on the floor and got 5.57m using 1/2*mv2 = ffd
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Then the problem asks, what will be the final speed at the bottom of the ramp if it has an initial speed of .6m/s?

and this is where I get stuck, I am not sure how to calculate that into the equation.

Thanks in advanced for the help.
 
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Welcome to PF :wink:
Use the template please.
Where's the figure?
 
thanks, Just updated the thread with the figure hopefully it works for you.
 
I didn't check your numbers, but with an initial velocity, the crate has initial kinetic energy. So use the same energy equation, and include the initial KE when calculating the kinetic energy change.
 
woww that totally worked, thanks a lot!
 
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