At which position is the LOWEST potential energy?
1 answer:
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Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
A) d = v² / (2g (μ cos θ + syn θ) B) μ = tan θ
Explanation:
Part A
We can work this part with the work and energy theorem, where the work of the friction forces is equal to the energy change of the system.
The work is
W = fr .d
With the force of friction it opposes the movement
W = - fr d
The energy at the lowest point is
Em₀ = K = ½ m v²
The energy at the highest point
= U = m g y
The height (y) can be found by trigonometry
sin θ = y / d
y = d sin θ
W = –Em₀
-fr d = mg d sin θ - ½ m v²
The equation for the force of friction is
fr = μ N
From Newton's second law
N - W cos Te = 0
We replace
-μ (mg cos θ) d - mg d sin θ = - ½ m v²
d g (μ cos θ + sin θ) = ½ v²
d = v² / (2g (μ cos θ + syn θ)
Part B
The block is stopped, what is the Angle tet, let's use Newton's second law
fr - W sin θ = 0 ⇒ fr = W sin θ
N - W cos θ= 0 ⇒ N = w cos θ
fr = μ N
μ (mg cos θ) = mg syn θ
μ = syn θ / cos θ
μ = tan θ