Dynamics And Kinematics Problem

  • #1
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Homework Statement


Given: m = 17kg Vi = 0 Fa=42N θ=25 Mu=0.15 T=125s

Diagram: http://s14.postimage.org/euy3whikh/New_Bitmap_Image.png [Broken]

Find: Δd

a block accellerates from rest for 125s find the displacement. Frictional forces act on the block.
A force applied is 25 degrees above the horizontal. Find the displacement

Homework Equations


Fnet=Ma
Δ=vit+a/2(t)^2

The Attempt at a Solution



Fnety=Fn + FaSin25 -Fg = 0

Fn= Fg - FaSin25

Fnetx=mAx=FaCos25 - Ffr

Ffr=Mu*Fn

Ffr= Mu*(Fg - FaSin25)

mAx = Facos25 - Mu*(Fg-FaSin25)

Ax= (Facos25 - Mu*(Fg-Fasin25))/m

Ax= 0.93 m/s^2 [--->]

Δd = vit + Ax/2(t)^2

Δd = + Ax/2(t)^2

answer: Δd = 7265.63 m

I'm not sure if this is correct or not, but my friend took a note in which it was the same question and he wrote down 78m.. I've done this question by myself to check everything but it bothers me the answer was wrong on the note he took, can some one infrom me if this is the correct way and if i have the correct answer to this problem please?

Did i not take something into account maybe..?
 
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Answers and Replies

  • #2
haruspex
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The only fault I can find in your working is that you cut Ax off at two decimal places then used it to get an answer you quote to 6. (7232 m is closer.)
For the block to travel only 78 m in 125 secs the acceleration would have to be tiny. Even if you neglect the sin theta term it still gives 6000m.
 
  • #3
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The only fault I can find in your working is that you cut Ax off at two decimal places then used it to get an answer you quote to 6. (7232 m is closer.)
For the block to travel only 78 m in 125 secs the acceleration would have to be tiny. Even if you neglect the sin theta term it still gives 6000m.

Thank you very much for checking this out for me i appreciate it very much. This settles my uneasiness about the question llol.
 

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