A crate is dragged 5.0 along a rough floor with a constant velocity by a worker applying a force of 125N to a rope at an angle of 30degrees to the horizontal.
Calculate the work done by the tension in the rope, the force of friction, and the net force. I know the work don on the weight of the crate and the normal force is 0J. Thanks for your help, and if possible show work, thanks!
Tension force in the direction of motion = 125 cos 30
= 108 N
Work done by rope tension =
Tension force in the direction of motion times distance moved = 108*5 = 541 Joules
Since the velocity is constant, the horizontal tension force, 108 N is equal and opposite to the friction force.
Therefore if the tension force is positive, the friction force is -108 N
The crate is not accelerating.
F = m a
a is zero
therefore
the net force on the crate is zero.
To calculate the work done by the tension in the rope, force of friction, and net force, we need to break down the forces acting on the crate.
1. Work done by the tension in the rope:
The work done by the tension in the rope can be calculated using the formula: Work = Force x Distance x cos(theta), where theta is the angle between the force and the displacement.
Given:
Force = 125N
Distance = 5.0m
Angle (theta) = 30 degrees
Work = 125N x 5.0m x cos(30 degrees)
Work = 125N x 5.0m x 0.866 (rounded to 3 decimal places)
Work = 541.25 Joules (rounded to 2 decimal places)
Therefore, the work done by the tension in the rope is 541.25 J.
2. Force of friction:
Since the crate is being dragged with a constant velocity, the net force acting on it must be zero. This means that the force of friction opposes the tension in the rope.
Therefore, the force of friction is equal in magnitude but opposite in direction to the tension force applied via the rope. Hence, the force of friction is 125N.
3. Net force:
The net force acting on the crate is the vector sum of all the forces acting on it. Since the crate is moving at a constant velocity, the net force must be zero. This means that the tension force and the friction force must cancel each other out.
Therefore, the net force on the crate is 0N.
To summarize:
- Work done by the tension in the rope = 541.25 J
- Force of friction = 125 N
- Net force = 0 N
To calculate the work done by the tension in the rope, the force of friction, and the net force, we need to first understand the different forces acting on the crate.
1. Work done by the tension in the rope:
Work done is equal to the force applied multiplied by the displacement in the direction of the force. In this case, the tension in the rope is the force applied, and the displacement is the distance the crate is dragged.
Work done by the tension = Force x displacement
= 125N x 5.0m
= 625 J (Joules)
So, the work done by the tension in the rope is 625 Joules.
2. Force of friction:
Since the crate is being dragged with a constant velocity, the net force acting on it must be zero. Therefore, the force of friction must be equal in magnitude to the applied force.
Force of friction = 125N
So, the force of friction is 125 Newtons.
3. Net force:
As mentioned earlier, since the crate is moving with a constant velocity, the net force must be zero. The net force is the vector sum of all the forces acting on the crate.
Net force = Force of tension + Force of friction
= 125N - 125N
= 0N
So, the net force is zero Newtons.
Overall, the work done by the tension in the rope is positive because it is in the same direction as the displacement. The work done by the force of friction is zero because it acts perpendicular to the displacement. And the net force is zero because the crate is moving at a constant velocity.