Calculate the work done by friction of package

Click For Summary
SUMMARY

The discussion focuses on calculating the work done by friction on a 5.00 kg package sliding down a 12-degree inclined ramp with a coefficient of kinetic friction of 0.310. The work done by friction is calculated as -22.2 J, while the work done by gravity is 15.3 J, resulting in a net work of -6.9 J. The final speed of the package, starting from an initial speed of 2.2 m/s, is determined to be 1.44 m/s. The calculations are confirmed as correct despite initial doubts about the necessity of breaking forces into components.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with work-energy principles
  • Knowledge of friction coefficients and their application
  • Basic trigonometry for resolving forces on inclined planes
NEXT STEPS
  • Study the concept of force resolution into x and y components
  • Learn about the work-energy theorem in physics
  • Explore the effects of different coefficients of friction on motion
  • Investigate the relationship between initial and final velocities in work-energy problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and tutors assisting with homework related to forces and motion on inclined planes.

veronicak5678
Messages
143
Reaction score
0

Homework Statement



A 5.00 kg package slides 1.50 m down a ramp that is inclined at 12 degrees below horizontal. The coefficient of kinetic friction is 0.310.
a- Calculate the work done by friction.
b- Calculate the work done by gravity.
c- Determine the net work done on package.
d- If thepackage started sliding with a speed of 2.2 m/s, find the final speed.

Homework Equations



Fk = n * coefficient
W = Fd cos angle
K= i/2 mv^2

The Attempt at a Solution



a-
n (Normal) = wcos 12= 47.9 N
Fk (Friction) = normal * coefficient = 14.8N
Work Fk = 1.5m * 14.8 N * cos 180 = -22.2 J

b-
wsin12 * 1.5m * cos 0 = 15.3 J

c-
W total = 15.3J - 22.2 J = -6.9 J

d-
1/2 m v init ^2 - 6.9 J = 1/2 m vfinal ^2

12.1 J - 6.9 J = 2.5 kg v final ^2

v final = 1.44 m/s


I was feeling pretty good about this, but I showed it to someone in my school's math lab and he told me he thought it was all wrong. He said I need to break every force in x and y components. I don't see why I would. I would appreciate it if someone could tell me if I am doing this correctly.
 
Physics news on Phys.org
You are correct.
 
Alright! Thank you.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 58 ·
2
Replies
58
Views
6K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
995
Replies
29
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 28 ·
Replies
28
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K