Speed of Block after Newton's Second Law

Click For Summary
SUMMARY

The discussion centers on applying Newton's Second Law and the kinematic equation v = v0 + at in a physics problem involving motion after an object starts moving. The user correctly identifies the need to calculate acceleration using the formula (Tcos∂ - ((mg - Tsin∂) * μk)) / m for part A. For part B, it is confirmed that since the object has just started moving, the initial velocity v0 is indeed 0, simplifying the equation to v = at.

PREREQUISITES
  • Understanding of Newton's Second Law
  • Familiarity with kinematic equations
  • Knowledge of static and kinetic friction coefficients
  • Basic algebra for solving equations
NEXT STEPS
  • Study the application of Newton's Second Law in various scenarios
  • Learn about the effects of static and kinetic friction on motion
  • Explore advanced kinematic equations and their derivations
  • Practice solving physics problems involving forces and motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion analysis, as well as educators looking for examples of applying Newton's laws in problem-solving contexts.

idllotsaroms
Messages
26
Reaction score
0

Homework Statement



http://postimage.org/image/6exf0w475/

B)
I have a question on this part. Would I just use the "v = v0 at" equation here because the question states "after it starts moving?" Do i need to include static/kinetic friction into this equation somehow?

Homework Equations



part a) Newton's Second Law

part b) v=v0 at

The Attempt at a Solution



For part A, I found the acceleration to be (Tcos∂-((mg-Tsin∂)*[itex]\mu[/itex]k)) / m

For part B I'm thinking that I use "v = v0 at" but the "after it starts moving" part of the question confuses me. Does it mean that v0 would be 0?
 
Physics news on Phys.org
idllotsaroms said:
For part A, I found the acceleration to be (Tcos∂-((mg-Tsin∂)*[itex]\mu[/itex]k)) / m
That's fine. (Where you use T instead of F.)

For part B I'm thinking that I use "v = v0 at" but the "after it starts moving" part of the question confuses me. Does it mean that v0 would be 0?
Yes, v0 = 0. The equation you mean is v = v0 + at.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
13
Views
3K
Replies
37
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
1K
Replies
16
Views
3K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K