A block is pushed 2m along a fixed horizontal surface by a horizontal force of 2N. The opposing force of friction is 0.4N. How much work is done by the frictional force
Well, well, well, looks like we have a frictional force stealing the show. Let's calculate the work done by this sneaky force.
Work is defined as the product of force and displacement. In this case, the opposing force of friction is 0.4N, and the block is pushed 2m. So, the work done by the frictional force is:
Work = Force × Displacement
Work = 0.4N × 2m
Work = 0.8 Joules
So, the frictional force does 0.8 Joules of work. It might not be putting on a grand performance, but hey, it's still getting the job done!
To find out how much work is done by the frictional force, we need to first determine the net force acting on the block.
The net force is the difference between the applied force and the opposing force of friction.
Net force = Applied force - Frictional force
Net force = 2N - 0.4N
Net force = 1.6N
The next step is to calculate the work done by the frictional force. Work is given by the equation:
Work = Force x Displacement x cos(θ)
In this case, the displacement is 2m, the force is the frictional force of 0.4N, and the angle between the force and displacement vectors is 0 degrees (cos(0) = 1).
Work = 0.4N x 2m x cos(0)
Work = 0.8J
Therefore, the frictional force does 0.8 joules of work.
To determine the work done by the frictional force, we need to know the displacement of the block and the magnitude of the force applied by friction. In this case, the displacement is given as 2m, and the magnitude of the opposing force of friction is 0.4N.
The work done by a force is calculated using the formula:
Work = Force x Displacement x cos(θ)
Where:
- Force is the magnitude of the force
- Displacement is the magnitude of the displacement
- θ is the angle between the direction of the force and the direction of displacement
In this case, the force of friction acts in the opposite direction to the displacement, so the angle between the force and displacement is 180 degrees.
Plugging in the values:
Work = 0.4N x 2m x cos(180°)
cos(180°) = -1
Work = 0.4N x 2m x -1
Work = -0.8 Joules
Therefore, the work done by the frictional force is -0.8 Joules.
Note: The negative sign indicates that the work done by the frictional force is in the opposite direction of the displacement, as the frictional force acts against the motion of the block.