Finding Work done without mass and friction force

In summary: For a body to slide down a slope, there must be a net force in that direction. Which force is that, and how is it related to the coefficient of friction? (Hint: Think trigonometry.)In summary, the problem statement involves a box being dragged down a sloped surface and reaching a certain speed at the base. The goal is to find the coefficient of friction between the box and the surface, the work done to move the box downward and the work done by the friction force, and the potential energy of the box. However, the problem statement is incomplete as it does not specify the force being used to drag the box down the slope, and there are errors in the equations used to solve the problem.
  • #1
Ummuali
2
0

Homework Statement


A box was dragged downward from the state of rest through an inclination 10.0 m at an angle of 25 degree. The speed when the box arrived at the base is 3 m/s.

a) what is the coefficient force between the box and the surface?
b) how many work is done to move the box downward and the work done by the friction force of the box?
c) calculate the potential energy of the box.

Homework Equations


1) F net + mg sin 25 (F applied) = µmg cos 25 (F friction)
F net = µmg cos 25 - mg sin 25
ma= µmg cos 25 - mg sin 25
a = µg cos 25 - g sin 25

2) V(final) ^2 = V(initial)^2 + 2ad

3) µ= F friction/ F Normal

The Attempt at a Solution


By using the equation nom 2, I found the acceleration is 0.45 m/s/s (I assumed that initial velocity is 0). Then, I used nom 1 equation and got µ= 0.56.

But I get stuck with the second question. I don't know if we can search the work done without knowing the mass and if I have to find the mass, I totally don't know how.

Please help me. Thank you.
 
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  • #2
Hello Ummuali, :welcome:

You want to check your answer to a) as well. When I try a = µg cos 25 - g sin 25 I don't get the acceleration you found (correctly).
Could youpost stepsin detail ?
Basically, the equation 1) you use isn't correct: there is an external force (the dragging) you left out ! So F net = µmg cos 25 - mg sin 25 isn't complete.
However, I then end up with two unknowns (the dragging force and µ). Is the problem statement complete ?

Also: in c) what exactly is being asked for ?

--
 
  • #3
Further to BvU's comments, what do you mean by "coefficient force"? Do you mean coefficient of (kinetic) friction, or frictional force?
 
  • #4
Owh, haha something could be wrong with my equ. However, this is my trial:

1) V(f)²= V(o)² + 2ad
3²= 0² + 2 (a)(10)
9= 20a
0.45 =a

So, a= 0.45 m/s/s

2)

μ = Ffriction/ Fnormal OR μ*Fnormal= Ffriction

Fnet in x-axis:
Ffriction - mg sin 25= ma
(Ffriction - mg sin 25) ÷ m = a

Fnet in y-axis:
Fnormal - mg cos 25 = ma
Fnormal - mg cos 25 = m (0) -since there is no acceleration in upward direction
Fnormal - mg cos 25 = 0

Subtitute these 3 equations;

(Ffriction - mg sin 25) ÷ m = a
( μ*Fnormal - mg sin 25) ÷ m = a
(μ*mg cos 25 - mg sin 25) ÷ m = a
μg cos 25 - g sin 25 = a -cut all m off
μ(9.8) (0.91) - (9.8) (0.42) = 0.45
μ 8.9 - 4.12 = 0.45
μ = (0.45 + 4.12) ÷ 8.9
μ=0.51

Value differ because of decimals. =)

The problem statement only give the value of angle, velocity when it reach the base of the inclined plane (which at first, it is in the state of rest), the displacement (how long does the box traveled through the plane). About question C, I'll try to confirm back.

Please correct me.
 
  • #5
Two problems with that.
As BvU already pointed out, the problem statement says the box is dragged down the slope, not that it slides down under gravity. This implies the presence of another force, but that force is not given and does not appear in your equations.
The other problem is your ##\Sigma F=ma## equation for movement parallel to the plane. Which way does it accelerate? Which force(s) act in that direction and which oppose it?
 

Related to Finding Work done without mass and friction force

1. How can work be calculated without knowing the mass and friction force?

In order to calculate work without knowing the mass and friction force, we can use the formula W = F x d, where W is work, F is the applied force, and d is the distance over which the force is applied. This formula only requires knowledge of the applied force and the distance, and does not factor in the mass or friction force.

2. Why is it important to consider work without mass and friction force?

Considering work without mass and friction force allows us to understand the basic principles of work and its relation to force and distance. It also helps us to isolate and understand the effects of different forces on work, without being influenced by other factors such as mass and friction.

3. Can work be done without any mass or friction force?

Yes, work can be done without any mass or friction force. For example, if a person pushes a wall with all their strength, the wall may not move but the person is still exerting force over a distance, thus work is still being done.

4. How does ignoring mass and friction force affect the accuracy of work calculations?

Ignoring mass and friction force can affect the accuracy of work calculations, as they both play a role in determining the amount of work done. However, in situations where the effects of mass and friction are minimal, ignoring them may not significantly affect the accuracy of the calculation.

5. What are some real-life examples of work being done without mass and friction force?

Some real-life examples of work being done without mass and friction force include pushing a wall or lifting an object without it moving, using a lever or pulley system, or walking on a level surface. In all of these cases, force is being applied over a distance, but the mass and friction force may not have a significant impact on the amount of work being done.

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