frog301

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May 25, 2018
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I do not understand how they got 'A' as the answer.

Here's how I initially tackled the problem, which turned out to be so unnecessarily confusing.

Solved for acceleration first (gsin30=5).
a=5
Solved for t going down ramp. 10=1/2at^2 -> t=sqrt(10(2)/a) -> t=sqrt(4)=2 seconds

Solved for Vf. Vf+Vi/2=d/t -> Vf=(10/2)(2) - 0 = 10 m/s

Now b/c we know regarding inclined plane, a=gsin(theta). At the bottom of the ramp, theta=0 or 180. sin0=0, meaning no change in velocity.

So with constant velocity at the bottom of the ramp, to travel distance 'x' the Vi=Vf, because it doesn't change due to 0 acceleration.

Vf+Vi/2=d/t. 10+10/2=10/t. t=10/10= 1 second to travel 'x' distance.

Did I do this problem correctly? I know I got the right answer, but unsure if my logic using constant velocity due to a=0 from gsin(0) was correct when I solved for time to travel 'x' distance.


*answer is A btw

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AdaptPrep

AdaptPrep MCAT
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Jan 19, 2017
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Hi, frog301--

Your logic is correct. Once the box hits the bottom and travels horizontally, you are correct that the velocity is constant (no acceleration since there is no incline and it is frictionless).

However, this might be easier to solve using conservation of energy.

You can calculate the distance of "h" as 5 m (sin 30 = h/10).

The potential energy of the box at "h":
PE = mgh = (mass)(10 m/s^2)(5 m)

The kinetic energy of the box after sliding down the ramp is:
KE = 1/2 (mass)(v^2)

Conservation of energy since it was a frictionless ramp; set them equal to each other:
PE = KE
(mass)(10 m/s^2)(5 m) = 1/2 (mass)(v^2)
50 = (1/2) (v^2)
100 = v^2
v = 10 m/s

Since it travels 10 m after hitting the bottom of the ramp:
(10 m) (1s/10m) = 1 second

Happy Studying!
 

frog301

2+ Year Member
May 25, 2018
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Hi, frog301--

Your logic is correct. Once the box hits the bottom and travels horizontally, you are correct that the velocity is constant (no acceleration since there is no incline and it is frictionless).

However, this might be easier to solve using conservation of energy.

You can calculate the distance of "h" as 5 m (sin 30 = h/10).

The potential energy of the box at "h":
PE = mgh = (mass)(10 m/s^2)(5 m)

The kinetic energy of the box after sliding down the ramp is:
KE = 1/2 (mass)(v^2)

Conservation of energy since it was a frictionless ramp; set them equal to each other:
PE = KE
(mass)(10 m/s^2)(5 m) = 1/2 (mass)(v^2)
50 = (1/2) (v^2)
100 = v^2
v = 10 m/s

Since it travels 10 m after hitting the bottom of the ramp:
(10 m) (1s/10m) = 1 second

Happy Studying!
Thank you!
 

PlsLetMeIn21

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Dec 5, 2017
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I did it a little differently but I think it's a good way to work it out. On the ramp it starts at 0 m/s and finishes at some v (which from your solution we know is 10). On the horizontal surface it maintains a constant speed of v (which is 10). So, the average speed across the horizontal surface is twice the average speed going down the ramp. x and d are both the same distance, so the time to travel the distance x is half of the time it takes to go down the ramp. For the ramp portion, d = 1/2at^2, where d is 10m and a is 5 m/s^2.

10 = 1/2(5)t^2 = 2.5t^2
4 = t^2
2 = t

It takes 2 seconds to go down the ramp, so it will only take 1 s to cover that same distance once it hits the bottom.
 
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