# Homework Help: Finding Work done without mass and friction force

1. Dec 13, 2015

### Ummuali

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

2. Relevant 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

3. 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.

2. Dec 13, 2015

### BvU

Hello Ummuali,

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. Dec 13, 2015

### haruspex

Further to BvU's comments, what do you mean by "coefficient force"? Do you mean coefficient of (kinetic) friction, or frictional force?

4. Dec 13, 2015

### Ummuali

Owh, haha something could be wrong with my equ. However, this is my trial:

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 travelled through the plane). About question C, I'll try to confirm back.

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?