# Work and Energy Block on incline problem

1. Oct 20, 2008

### Thepoint

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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.

2. Oct 20, 2008

### Perillux

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.

3. Oct 20, 2008

### LowlyPion

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?

4. Oct 20, 2008

### Perillux

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.

5. Oct 20, 2008

### Thepoint

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:)

6. Nov 7, 2008

### ionmorris

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)