A man drags a sack of flour of mass 50 kg at
constant speed up an inclined plan to a height of 5m. The plane makes an angle of 30„awith the horizontal and a constant frictional force of 200 N acts on the sack down the plane.
1- How much work does the man do
A- against friction.
B- against gravity.
C- against normal force.
2- Total work done by the man.
To determine the work done by the man in each scenario, we need to understand the definition of work and the forces involved. Work is given by the equation W = Fd cosθ, where W is work, F is the force applied, d is the displacement, and θ is the angle between the force and the direction of displacement.
1. Work done against friction:
We know that the sack is being dragged up the inclined plane. The frictional force acts down the plane, opposing the motion. Since the sack is moving at a constant speed, the force exerted by the man must be equal in magnitude but opposite in direction to the frictional force.
The work done against friction can be calculated by multiplying the force of friction by the displacement of the sack:
Work against friction = Frictional force x displacement x cosθ
In this case:
Frictional force = 200 N (given)
Displacement = 5 m (given)
θ = angle between the force of friction and the direction of displacement, which is 180° as the force and displacement are in opposite directions.
Plugging in the values:
Work against friction = 200 N x 5 m x cos 180°
= -1000 J
Note that the negative value implies that work is done in the opposite direction of displacement.
2. Work done against gravity:
The man is lifting the sack vertically against gravity, which means he is doing work against the gravitational force acting on the sack.
The work done against gravity can be calculated by multiplying the weight of the sack by the vertical displacement:
Work against gravity = Weight x displacement x cosθ
Weight = mass x gravitational acceleration
= 50 kg x 9.8 m/s² (acceleration due to gravity)
= 490 N
Displacement = 5 m (given)
θ = angle between the force of gravity and the direction of displacement, which is 90° since the force of gravity is acting downward perpendicular to the inclined plane.
Plugging in the values:
Work against gravity = 490 N x 5 m x cos 90°
= 0 J
Note that the work against gravity is zero since the force of gravity is perpendicular to the displacement.
3. Work done against normal force:
The normal force acts perpendicular to the inclined plane. Since the man is moving the sack along the inclined plane, no work is done against the normal force. Therefore, the work done against the normal force is zero.
4. Total work done by the man:
To calculate the total work done, we add up the work done against friction, gravity, and normal force since they are the only forces involved:
Total work done = Work against friction + Work against gravity + Work against normal force
= -1000 J + 0 J + 0 J
= -1000 J
The total work done by the man is -1000 Joules. The negative sign represents that the work done by the man is in the opposite direction of the displacement.
You should be able to get some or all of these.
They tell you what the friction force is. Multiply that by the distance he moves to get work against friction.
They work against gravity is M g H, where H is the height change.
There can be no work done against a force that is perpendicular to the direction of motion.
That information should help you come up with the answers yourself