D’Alembert Vs Conservation of Energy

AI Thread Summary
The discussion revolves around solving a physics problem involving a block being pulled up an incline, requiring the application of D’Alembert’s principle and the conservation of energy. The calculations include determining the frictional force, net force, work done, and power, while also addressing the need to account for energy lost due to friction. An arithmetic error in the acceleration calculation was pointed out, and it was suggested that using conservation of energy might complicate the solution due to friction being a non-conservative force. The user confirmed the necessity of employing both principles for their assignment and sought validation of their equations. Ultimately, the numerical answers were confirmed as correct, providing a basis for comparison with other methods.
GrimUpNorth
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Homework Statement


A block with a mass of 5kg is pulled up an incline with a 20 degree incline, the coefficient of friction is 0.2 and the distance 10 metres. The block accelerates from 1m/s at point A to 5m/s at point B.
I have to use both D’Alembert’s and Conservation of energy principles.

Does the following look correct? I only have one chance to get it right so want to make sure it’s correct.

Homework Equations


Frictional force=µ*M*g*cosӨ
a=(v2-u2)/2s
Net force=Ma
t=(v-u)/a

W + PE(initial) +KE(initial) =PE(final) +KE(final) +Heat lost due to friction
3. The Attempt at a Solution [/B]
Mass of the block=5kg
Angle of inclination=20 ̊
Displacement, s=10m
Initial velocity=1m/s
Final velocity=5m/s
Coefficient of friction, µ=0.2
From Alembert’s principle, net force-Ma=0 (Ghosh, 2016, pp.9)
Net force=pulling force -frictional force
Frictional force=µ*M*g*cosӨ
=0.2*5*9.81*cos (20)
=9.218 N
From Newtons laws, v2=u2+2as
a=(v2-u2)/2s
= (25-1)/2*10
=2.4m/s2
Therefore, Ma=5*2.4=12N
From the above principle,
Net force=Ma
=12N
Pulling force=Net force + frictional force
=12N + 9.218N
=21.218N
Work done in moving the block from point A to B equals pulling force*displacement(AB)
=21.218*10
=212.18J
From Newtons laws=U+ at
Therefore, t=(v-u)/a
= (5-1)/2.4
=1.66sec
Power=work done/time
=212.18/1.66 =128.2 Watts
Principle of conservation of energy
From this principle, energy can only be transformed into other forms but cannot be created or destroyed. Therefore, the net work done by the forces is zero as mechanical energy is conserved.
Work input=W
potential energy=PE
Kinetic energy =KE
Using this principle, we obtain the following equation;
W + PE(initial) +KE(initial) =PE(final) +KE(final) +Heat lost due to friction……(i)
PE(initial)=0, since initial height is zero
KE(initial)=(m*u2)/2= 2.5J
PE(final) =(mgh)=5*9.81*cos (20)
=46.1J
KE(final)=(m*v2)/2= (5*52)/2
=62.5J
Heat lost due to friction=µMg*cosӨ*d=0.2*5*9.81*cos20*10
=92.18J
Equation (i) above reduces to:
W+KE=PE +KE+Energy lost as heat, since initial PE equals zero
Substituting the corresponding values,
W+2.5=46.1+62.5+92.18
Making W the subject of the formula,
W=-2.5+46.1+62.5+92.18
=198.28J
From work done obtained above(W=198.28J), pulling force will be
F=w/d
=198.28/10=19.828N
 
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Though you do not say so, I assume you are trying to solve for the force applied to pull the block up the ramp.

You have an arithmetic error here:
GrimUpNorth said:
a=(v2-u2)/2s
= (25-1)/2*10
=2.4m/s2

I would not use conservation of energy here since friction is a non-conservative force. You do account for the heat lost to friction, but I think you are making things unnecessarily complicated. I suggest that you write the equation for the net acceleration up the ramp, including the pulling force, friction and component of the force of gravity pulling down the ramp. You can also find the net acceleration in terms of the distance traveled and initial and final velocities of the block. Set the two expressions for net acceleration equal and solve for the pulling force.
 
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tnich said:
Though you do not say so, I assume you are trying to solve for the force applied to pull the block up the ramp.

You have an arithmetic error here:I would not use conservation of energy here since friction is a non-conservative force. You do account for the heat lost to friction, but I think you are making things unnecessarily complicated. I suggest that you write the equation for the net acceleration up the ramp, including the pulling force, friction and component of the force of gravity pulling down the ramp. You can also find the net acceleration in terms of the distance traveled and initial and final velocities of the block. Set the two expressions for net acceleration equal and solve for the pulling force.
Thank you, I have to use both principles and then compare and contrast the results for the assignment.
tnich said:
Though you do not say so, I assume you are trying to solve for the force applied to pull the block up the ramp.

You have an arithmetic error here:I would not use conservation of energy here since friction is a non-conservative force. You do account for the heat lost to friction, but I think you are making things unnecessarily complicated. I suggest that you write the equation for the net acceleration up the ramp, including the pulling force, friction and component of the force of gravity pulling down the ramp. You can also find the net acceleration in terms of the distance traveled and initial and final velocities of the block. Set the two expressions for net acceleration equal and solve for the pulling force.

Thank you, I need to use both principles for the assignment.
Does this look a better equation to use?
 

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GrimUpNorth said:
Thank you, I have to use both principles and then compare and contrast the results for the assignment.Thank you, I need to use both principles for the assignment.
Does this look a better equation to use?
Yes, I think you have the right numerical answer now. That will give you something to check the other methods against.
 
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