Finding Work done without mass and friction force

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The discussion focuses on solving a physics problem involving a box dragged down an incline. The key equations used include net force calculations and kinematic equations to find acceleration and the coefficient of friction. There is confusion regarding the presence of an external force in the equations, as well as the interpretation of terms like "coefficient force." Participants highlight the need for clarity in the problem statement and the correct application of forces acting on the box. The conversation emphasizes the importance of accurately identifying all forces involved to solve for work done and potential energy.
Ummuali
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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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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 ?

--
 
Further to BvU's comments, what do you mean by "coefficient force"? Do you mean coefficient of (kinetic) friction, or frictional force?
 
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.
 
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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