Work and Energy Block on incline problem

AI Thread Summary
The discussion revolves around calculating the final velocity of a box being pushed up a frictionless incline using work-energy principles. The user initially calculates the net force by subtracting the gravitational component from the applied force, leading to an acceleration of 3.14 m/s². However, confusion arises regarding the final velocity, with a calculated value of 7.09 m/s, which may be incorrect due to significant figure considerations or misunderstanding the problem's requirements. The correct approach involves determining the net work done and applying it to find the kinetic energy, ultimately leading to the conclusion that the expected answer might be around 5.5 m/s. Clarifications on significant figures and answer formats are suggested for resolving discrepancies in calculated results.
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)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top