(a)Calculate the work done by moving a 50 kg mass along each of the legs of the path shown below. What is the total work expended when going along the entire path? Show all work. Don't just state any obvious results, if any.
AB= 5m
BC= 10m
CD= 19.14m
DA= 10m
Angle between AB= 45 degrees,
angle between b and c equals 45 degrees.
(b) Do part a when the path ABCD lies on a level floor with coefficient of kinetic friction µk= 0.15 Show all work.
(a) To calculate the work done along each leg of the path, we need to use the formula:
Work = Force × Distance × Cosine(theta)
where Force is the force applied to move the mass, Distance is the distance traveled along the leg, and theta is the angle between the force and the displacement.
Let's calculate the work done along each leg:
1. Leg AB:
The distance traveled along AB is 5m, and the angle between the force and the displacement is 45 degrees. We need to find the force applied to move the mass.
Since the mass is not given an acceleration, we can assume that it is being moved at a constant velocity. Therefore, the net force acting on the mass is zero. The only force acting against the motion is the force of gravity, given by F = m × g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, the force applied to move the mass along AB is equal in magnitude but opposite in direction to the force of gravity, which is F = 50 kg × 9.8 m/s^2.
Now we can calculate the work done along AB:
Work_AB = Force × Distance × Cosine(45 degrees) = - (50 kg × 9.8 m/s^2) × 5 m × Cosine(45 degrees)
2. Leg BC:
The distance traveled along BC is 10m, and the angle between the force and the displacement is also 45 degrees.
Using the same reasoning as in leg AB, the force applied to move the mass along BC is also - (50 kg × 9.8 m/s^2).
Now we can calculate the work done along BC:
Work_BC = Force × Distance × Cosine(45 degrees) = - (50 kg × 9.8 m/s^2) × 10 m × Cosine(45 degrees)
3. Leg CD:
The distance traveled along CD is 19.14m, and the angle between the force and the displacement is not given. Since the angle is not provided, we can assume that the force applied to move the mass is perpendicular to the displacement, resulting in zero work done along this leg.
Therefore, Work_CD = 0
4. Leg DA:
The distance traveled along DA is 10m, and the angle between the force and the displacement is also not given. Similar to leg CD, we can assume that the force applied to move the mass is perpendicular to the displacement, resulting in zero work done along this leg.
Therefore, Work_DA = 0
Now, to find the total work expended when going along the entire path, we just need to sum up the work done along each leg:
Total Work = Work_AB + Work_BC + Work_CD + Work_DA
(b) When the path ABCD lies on a level floor with a coefficient of kinetic friction µk = 0.15, the work done will differ from part (a) due to the presence of friction.
Friction opposes the motion and is given by the equation Friction = µk × Normal Force.
To find the normal force, we need to consider the force balance in the vertical direction. Since the mass is on a level floor, the normal force is equal in magnitude but opposite in direction to the force of gravity, given by Normal Force = m × g.
Therefore, the frictional force acting against the motion is given by Friction = µk × (m × g).
Using the work formula mentioned earlier, the work done along each leg can now be calculated as follows:
1. Leg AB:
The force applied to move the mass along AB is equal in magnitude but opposite in direction to (m × g + Friction). The distance traveled along AB is 5m, and the angle between the force and the displacement is 45 degrees.
Now we can calculate the work done along AB:
Work_AB = (-(50 kg × 9.8 m/s^2 + µk × (50 kg × 9.8 m/s^2))) × 5 m × Cosine(45 degrees)
2. Leg BC:
Similar to leg AB, the force applied to move the mass along BC is equal in magnitude but opposite in direction to (m × g + Friction). The distance traveled along BC is 10m, and the angle between the force and the displacement is 45 degrees.
Now we can calculate the work done along BC:
Work_BC = (-(50 kg × 9.8 m/s^2 + µk × (50 kg × 9.8 m/s^2))) × 10 m × Cosine(45 degrees)
The work done along legs CD and DA will still be zero since the force is assumed to be perpendicular to the displacement.
Total Work = Work_AB + Work_BC + Work_CD + Work_DA
This calculation considers the presence of friction and provides a more accurate representation of the work expended when moving along the given path on a level floor.