a man pulls a log of mass 45 kg along level ground by means of a rope at constant speed. If the applied force is at an angle of 30 degrees to the horizontal and the force of friction is 36N find the work done by him pulling the log through a distance of 10m
To find the work done by the man pulling the log, we need to calculate the work done against friction and the work done due to the applied force.
1. Work done against friction:
The force of friction acts opposite to the direction of motion, so the work done against friction can be calculated using the formula: Work against friction = Force of friction * distance
Given that the force of friction is 36N and the distance is 10m, we can substitute these values into the formula:
Work against friction = 36N * 10m = 360 Joules
2. Work done due to the applied force:
The applied force can be divided into two components: the horizontal component and the vertical component. The work done by the vertical component is zero because the displacement is along level ground (perpendicular to the vertical).
The work done by the horizontal component can be calculated using the formula: Work due to applied force = Applied force * distance * cos(theta)
Given that the mass of the log is 45 kg and the angle with the horizontal is 30 degrees, we can calculate the applied force using the formula: Applied force = mass * acceleration due to gravity
Applied force = 45 kg * 9.8 m/s^2 = 441 N
Now, we can substitute these values into the formula:
Work due to applied force = 441 N * 10m * cos(30 degrees) = 441 N * 10m * cos(π/6)
Work due to applied force = 441 N * 10m * (√3/2) = 2205√3 Joules
To find the total work done, we can sum up the work done against friction and the work done due to the applied force:
Total work done = Work against friction + Work due to applied force
Total work done = 360 Joules + 2205√3 Joules ≈ 2182.2 Joules
Therefore, the man does around 2182.2 Joules of work by pulling the log through a distance of 10m along level ground.