Inclined plane and elapsed time

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Homework Help Overview

The problem involves a block moving up an inclined plane with an initial speed, requiring calculations for the distance it travels up the incline and the time taken to return to the starting point. The subject area includes kinematics and the application of equations of motion along an incline.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the time and distance traveled by the block. There is a focus on the correct application of acceleration components along the incline and the need to clarify the signs of acceleration. Some participants question the arithmetic and the setup of the equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and questioning the assumptions made regarding acceleration. There is no explicit consensus on the correct approach yet, as multiple interpretations of the equations and their components are being explored.

Contextual Notes

Participants note potential errors in arithmetic and the need to reconsider the signs of acceleration due to the incline's direction. There is also a mention of the initial speed being parallel to the incline, which raises questions about how to apply trigonometric functions in the calculations.

Sassenav22
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Homework Statement


A block is given a initial speed of 5.0 m/s up the 20° plane. (a) How far up the plane will it go?
(b) how much much time elapses before it returns to its starting point?


Homework Equations


I think

V= u + at
s=ut t 1/2at^2
v=d/t

The Attempt at a Solution



(a) V= u + at
0= 5-9.8(sin20)t
t = 1.65s

s=ut+1/2at^2
=5(sin20)(1.65)+ 1/2(9.8)(1.65^2)
s= 16.16m

(b) v=d/t
t=d/v
=16.16/5
t= 3.2s


I know this is wrong but i can't figure it out can someone help me?
 
Last edited:
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Sassenav22 said:

The Attempt at a Solution



V= u + at
0= 5-9.8(sin20)t
t = 1.65s
Recheck your arithmetic. (Tip: Once you've solved for the time the block takes to rise, what's the answer to part b?)

s=ut+1/2at^2
=5(sin20)(1.65)+ 1/2(9.8)(1.65^2)
s= 16.16m
The initial speed up the ramp is already parallel to the incline, so there's no need to multiply it by sin20. On the other hand, the acceleration along the incline is only a component of g, so you do need the sin20 there. Also, should the acceleration be + or -?
 
acceleration should be positive
 
So what your saying is that the sin20 is suppose to beby the 9.8?
 
Sassenav22 said:
acceleration should be positive
No. The ramp slopes up, so acceleration is negative (down).
Sassenav22 said:
So what your saying is that the sin20 is suppose to beby the 9.8?
Yes, the acceleration of the block is g sin20.
 
Oh, Is the first part of (a) correct because I found the time so that i could find the distance, is that what is suppose to be done?
 
Sassenav22 said:
Oh, Is the first part of (a) correct because I found the time so that i could find the distance, is that what is suppose to be done?
Yes, that's a perfectly fine way to solve the problem. You just made an error when you did the final arithmetic to solve for the time.
 
so its:

(a) V=u+at
0=5+9.8(sin20)t
t = 1.65s

s= ut+1/2at^2
= 5(1.65)+1/2(-9.8)(sin20)(1.65^2)
=3.69m

(b) v=d/t
5=3.69t
t=3.39/5
t=0.68s
 
Sassenav22 said:
(a) V=u+at
0=5+9.8(sin20)t
t = 1.65s
The final answer is still wrong. Redo the calculation.

s= ut+1/2at^2
= 5(1.65)+1/2(-9.8)(sin20)(1.65^2)
=3.69m
You'll need to redo this with the correct time.

(b) v=d/t
5=3.69t
t=3.39/5
t=0.68s
No. Read my "tip" in post #2.
 
  • #10
v=u+at
= 5-9.8(sin20)t
t= 1.64s

I still get the same answer
 
  • #11
Sassenav22 said:
v=u+at
= 5-9.8(sin20)t
t= 1.64s

I still get the same answer
I'm curious how. t = 5/(9.8sin20) = ??
 

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