The length of the plane is 10 m. Let the coefficient of kinetic friction between the block and the plane be 0.2.
Obtain the acceleration of the block as it moves down the plane.
Determine the work done by the frictional force.
Calculate the total mechanical energy of the block at the bottom of the plane.
Obtain the acceleration of the block as it moves down the plane.
Determine the work done by the frictional force.
Calculate the total mechanical energy of the block at the bottom of the plane.
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A block of weight 98 N slides down a plane inclined at 45° to the horizontal from rest.?
Mass of block = 98 ÷ 9.8 = 10 kg
The length of the plane is 10 m. Let the coefficient of kinetic friction between the block and the plane be 0.2.
Obtain the acceleration of the block as it moves down the plane.
Net force = mass * acceleration
Force parallel = m * g * sin θ
Force of friction = µ * m * g * cos θ
Net force = m * g * sin θ – µ * m * g * cos θ
m * g * sin θ – µ * m * g * cos θ = m * a
Divide both sides m
g * sin θ – µ * g * cos θ = a
9.8 * sin 45 – 0.2 * 9.8 * cos 45 = acceleration = 5.5437 m/s^2
Determine the work done by the frictional force = Friction force * length of plane
work done by the frictional force = 0.2 * 10 * 9.8 * cos 45 * 10 = 138.6 J
Calculate the total mechanical energy of the block at the bottom of the plane.
The length of the plane is 10 m.
As the block moves down the incline, the potential energy decreases and the kinetic energy increases.
PE = m * g * height
Height of the incline = Length * sin θ = 10 * sin 45
Initial PE = 10 * 9.8 * 10 * sin 45 = 693 J
This is total energy at the top of the inclined plane.
Since there is friction, the kinetic energy at the bottom of the inclined plane = total energy at the top minus work done by the frictional force
kinetic energy at the bottom = 693 – 138.6 = 554.4 J
The total mechanical energy of the block at the bottom of the plane ≈ 554.4 J
OR
KE = ½ * m * v^2
Determine the final velocity^2 and final KE
vf^2 – vi^2 = 2 * a * d
vi = 0 m/s
a = 5.5437 m/s^2
d = 10 m
vf^2 = 2 * 5.5437 * 10 = 110.874
KE = ½ * 10 * 110.874
Final KE = 554.37 J
Mass of block = 98 ÷ 9.8 = 10 kg
The length of the plane is 10 m. Let the coefficient of kinetic friction between the block and the plane be 0.2.
Obtain the acceleration of the block as it moves down the plane.
Net force = mass * acceleration
Force parallel = m * g * sin θ
Force of friction = µ * m * g * cos θ
Net force = m * g * sin θ – µ * m * g * cos θ
m * g * sin θ – µ * m * g * cos θ = m * a
Divide both sides m
g * sin θ – µ * g * cos θ = a
9.8 * sin 45 – 0.2 * 9.8 * cos 45 = acceleration = 5.5437 m/s^2
Determine the work done by the frictional force = Friction force * length of plane
work done by the frictional force = 0.2 * 10 * 9.8 * cos 45 * 10 = 138.6 J
Calculate the total mechanical energy of the block at the bottom of the plane.
The length of the plane is 10 m.
As the block moves down the incline, the potential energy decreases and the kinetic energy increases.
PE = m * g * height
Height of the incline = Length * sin θ = 10 * sin 45
Initial PE = 10 * 9.8 * 10 * sin 45 = 693 J
This is total energy at the top of the inclined plane.
Since there is friction, the kinetic energy at the bottom of the inclined plane = total energy at the top minus work done by the frictional force
kinetic energy at the bottom = 693 – 138.6 = 554.4 J
The total mechanical energy of the block at the bottom of the plane ≈ 554.4 J
OR
KE = ½ * m * v^2
Determine the final velocity^2 and final KE
vf^2 – vi^2 = 2 * a * d
vi = 0 m/s
a = 5.5437 m/s^2
d = 10 m
vf^2 = 2 * 5.5437 * 10 = 110.874
KE = ½ * 10 * 110.874
Final KE = 554.37 J