Block Sliding Down A Ramp

  • #1
Hi! This is my first post, so if I do anything incorrectly I apologize :redface:


Homework Statement



A block of wood slides first down and then up a ramp inclined at 25°. Assume the block/ramp kinetic friction coefficient = 0.2. The block starts from rest at position located 3.2 m up from the lowest point on the ramp. After sliding down, the block bounces back at the lowest point. The speed just after the bounce is only 87% of what it was just before the bounce.

a) Draw the free-body diagram for the block during the downward motion.
b) Using the above diagram, find the block's acceleration on the way down.
c) How much time does the block travel until it hits the lowest point?
d) Draw the free-body diagram for the block during the upward motion.
e) Using this diagra, find the block's acceleration on the way up.
f) Find the time the block spends moving up after the bounce until it reaches the highest point.
g) How far is that point from the bounce-point?



Homework Equations


--

The Attempt at a Solution



a) I have the FBD drawn correctly, no help needed here.

b) a = -g sin(25°) - 0.2cos(25°)
a = -2.37 m/s^2 (I am pretty sure this is correct as well)

c) time = square root of 2d/a = square root of (2(3.2)/2.37)
t = 1.64 seconds (not feeling good about this part)

d) No help needed on the FBD.

e) -g sin(25°) + 0.2cos(25°)
a = -5.92 m/s^2

f) velocity = t * |a| = 1.64 * 2.37 = 3.89 m/s
3.89 * 82% = 3.19 m/s
t = v/|a| = 3.19/5.92 = 0.54 seconds

g) distance = v*t
d = (3.19)(0.54) = 1.72 m

I don't feel like I did c, f, or g correctly. Sorry this was so long and probably hard to read. I am just getting used to this! Any help would be appreciated, even if it is only for one of the letters. Thank you!
 

Answers and Replies

  • #2
kuruman
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Hi justbreathe87. Welcome to PF.

a) I have the FBD drawn correctly, no help needed here.

b) a = -g sin(25°) - 0.2cos(25°)
a = -2.37 m/s^2 (I am pretty sure this is correct as well)
It is not correct for the following reasons:

1. It is dimensionally incorrect. The first term has (correctly) dimensions of acceleration, but the second term does not.

2. When the block is sliding down the incline, the component of the acceleration g sin(25o) is in the direction of motion (down the incline) but the second term which is due to friction is in the opposite direction (up the incline). The two terms then must have opposite algebraic signs. If one is positive, the other one must be negative. You have written them both with the same sign.

Once you have the correct value for the acceleration you can proceed to the other parts.

Part (c) the method is correct.
Part (e), again you need to watch your dimensions and the relative signs. Here the two terms of the acceleration must have the same relative sign, because they both point down the incline when the block is moving up the incline.
Part (f), the method is correct
Part (g), you need to use the kinematic equation to find the distance. You know the initial velocity, you know the acceleration and you know the time. The equation
distance = v*t is valid only when the acceleration is zero.
 
Last edited:

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