A block is pushed 2 m along a fixed horizontal surface by a horizontal force of 10 N. The opposing force of friction is 0.4 N
(a)How much work is done by the 10 N force?
(b)what is the work done by the friction force
work= force.distance= 10*2=20n
friction=force.distance=.4*10*cos180=
remember the dot product means times the cosine of the angle between force and distance. In frictions case, the force opposes the displacement (180 degrees)
To answer part (a) of the question, you need to calculate the work done by the 10 N force. Work can be calculated using the equation:
Work = Force * Distance * Cosine(theta)
Where:
- Force is the magnitude of the force applied (10 N in this case)
- Distance is the displacement of the object (2 m in this case)
- Cosine(theta) is the angle between the direction of the force and the direction of displacement. Since the force and displacement are both horizontal, the angle is zero degrees and its cosine is 1.
Calculating the work done by the 10 N force:
Work = 10 N * 2 m * Cosine(0°)
= 10 N * 2 m * 1
= 20 joules
Therefore, the work done by the 10 N force is 20 joules.
To answer part (b) of the question, you need to calculate the work done by the friction force. The friction force opposes the motion of the block, so it does negative work.
The work done by the friction force can be calculated in the same way as before, but with the magnitude of the friction force (0.4 N) substituted for the applied force:
Work = Friction Force * Distance * Cosine(theta)
Since the friction force is acting in the opposite direction of the displacement, the angle between them is 180 degrees. The cosine of 180 degrees is -1.
Calculating the work done by the friction force:
Work = 0.4 N * 2 m * Cosine(180°)
= 0.4 N * 2 m * (-1)
= -0.8 joules
Therefore, the work done by the friction force is -0.8 joules.