Work and Energy Block on incline problem

In summary, a box is moved up an incline with a slope of 32° using a horizontal force of 150 N. The box's potential energy is increased by the work done against the gravitational force, and the box's kinetic energy is also increased by the work done moving it up the incline.
  • #1
Thepoint
4
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.
 
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  • #2
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
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?
 
  • #4
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
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
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)
 

1. What is the relationship between work and energy in a block on an incline problem?

In a block on an incline problem, work and energy are related through the principle of conservation of energy. This means that the work done on the block by external forces (such as gravity and friction) is equal to the change in the block's energy (kinetic and potential). This relationship can be expressed mathematically as W = ΔE, where W is work, and ΔE is the change in energy.

2. How do you calculate the work done on a block on an incline?

The work done on a block on an incline can be calculated using the equation W = Fcosθd, where F is the force applied on the block, θ is the angle between the force and the displacement, and d is the displacement of the block. This equation takes into account the component of the force that is parallel to the displacement, as only this component does work on the block.

3. What factors affect the work and energy in a block on an incline problem?

The work and energy in a block on an incline problem are affected by factors such as the applied force, the angle of the incline, the mass of the block, and the presence of friction. These factors can impact the amount of work done on the block and the total energy of the system.

4. How does the angle of the incline affect the work and energy in a block on an incline problem?

The angle of the incline affects the work and energy in a block on an incline problem by changing the component of the force that is parallel to the displacement. As the angle increases, the component of the force decreases, resulting in less work being done on the block. This also affects the potential energy of the block, as it is directly proportional to the height of the incline.

5. How does friction impact the work and energy in a block on an incline problem?

Friction impacts the work and energy in a block on an incline problem by dissipating some of the energy into heat and reducing the overall efficiency of the system. Frictional forces act in the opposite direction of motion, so they can decrease the amount of work done on the block and affect its kinetic and potential energy. Therefore, it is important to consider friction when analyzing a block on an incline problem.

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