Work and Energy Block on incline problem

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving work and energy, specifically analyzing the motion of a box being pushed up a frictionless incline. The problem includes calculating the final velocity of the box after being pushed a certain distance against gravitational forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of work-energy principles, including net work and kinetic energy changes. There are attempts to calculate the net force acting on the box and its resulting acceleration. Some participants question the accuracy of their calculations and the significance of significant figures in their answers.

Discussion Status

Several participants have provided different approaches to the problem, with some suggesting the use of work equations and others expressing uncertainty about their results. There is acknowledgment of potential discrepancies in answers due to significant figures, and one participant mentions receiving feedback from an interactive website regarding their answer.

Contextual Notes

Participants note that the problem involves a frictionless incline and that the calculations must consider both the applied force and gravitational work. There is also mention of the need to adhere to specific answer formats required by external programs.

Thepoint
Messages
4
Reaction score
0

Homework Statement


A horizontal force of magnitude F = 150 N is used to push a box of mass m = 18 kg from rest a distance d = 8 m up a frictionless incline with a slope q = 32°.

a, b. and c I already have done

d) How fast is the box moving after this displacement? [Hint: Work-energy involves net work done.] v = m/s


Homework Equations





The Attempt at a Solution


the pushing force is countered by the factor of gravitational work
the component force is 18*9.8*sin32=93.47778
F=150N, so net force pushs the object is 150-93.47778=56.52224N
net force=ma
m=18, net force =56.52224 so a=3.14m/s^2
Vf^2=2as, s=8m
Vf=7.08815m/s

I have no clue what I did wrong. Any assistance would be great.
 
Physics news on Phys.org
Based on the hint given in part d, I would assume that you should use the equations:
W = F*d
and
Wnet = (delta)KE (net work equals change in kinetic energy)

find the first one, and then use that value for Wnet in the second.
 
Thepoint said:

Homework Statement


A horizontal force of magnitude F = 150 N is used to push a box of mass m = 18 kg from rest a distance d = 8 m up a frictionless incline with a slope q = 32°.

a, b. and c I already have done

d) How fast is the box moving after this displacement? [Hint: Work-energy involves net work done.] v = m/s

Homework Equations



The Attempt at a Solution


the pushing force is countered by the factor of gravitational work
the component force is 18*9.8*sin32=93.47778
F=150N, so net force pushs the object is 150-93.47778=56.52224N
net force=ma
m=18, net force =56.52224 so a=3.14m/s^2
Vf^2=2as, s=8m
Vf=7.08815m/s

I have no clue what I did wrong. Any assistance would be great.

Increase in Potential energy is m*g*h = d*sinθ *m*g = 8*.53*9.8*18 = 747.94 N-m
Work from 150N over 8m = 1200 N-m
Net to Kinetic energy = 1200 - 747.94 = 452.06 = 1/2*m*V2
V2 = (2*452.06/18)1/2 = 7.09m/s

Looks like your answer method arrives at the right answer as well. I can only suggest it's a significant digit issue. All variables were given with no digits after the decimal, so perhaps the expected answer is merely 7 m/s?
 
how do you know you are wrong? If you are trying to enter the answer into a program you should read the answer requirements, usually those programs will tell you. Some like exact answers to a certain amount of digits, some account for slightly different answers, and some want significant figures to be used.
If the program doesn't tell you, then perhaps you can google it and find out what it expects for answers.
 
The actual answer to the question is 5.5m/s but I just guessed it. BTW I was using an interactive website so it told me my answer was wrong. Thanks for trying to help:)
 
so the Work total = W applied (answer of a) + W gravity(answer of b)
W total = F * D
F=W/D
F/m=a

Vf^2 = Vo^2 + 2a (D)
Vf^2 = 0 + 2a (D)
Vf = sqr root of (2a*D)
 

Similar threads

Replies
2
Views
851
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
10
Views
2K
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K