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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