(Please put the resulting figures and Thanks))))
a man pushes a box of mass 5 kg in a horizontal direction by 10 N force on the surface through a distance of 5m if the cofficien of kinetic friction is 0.2 the total work done on the box is
10 N x 5 m = ____ J
You do not need to know the friction coefficient or the mass, although they do affect the acceleration rate..
You do the figures
To find the total work done on the box, we need to consider both the work done by the applied force and the work done by the friction force.
1. Work done by the applied force:
The work done by a force is calculated using the formula:
work = force × distance × cos(angle)
In this case, the force applied by the man is 10 N, the distance is 5 m, and the angle between the force and the displacement is 0° (since the force is applied horizontally).
So, the work done by the applied force is:
work = 10 N × 5 m × cos(0°) = 50 joules
2. Work done by the friction force:
The friction force is equal to the coefficient of kinetic friction (μ) multiplied by the normal force (N).
The normal force can be calculated using the formula:
normal force = mass × gravitational acceleration
In this case, the mass of the box is 5 kg, and the gravitational acceleration is approximately 9.8 m/s^2.
So, the normal force is:
normal force = 5 kg × 9.8 m/s^2 = 49 N
The work done by the friction force is given by the formula:
work = friction force × distance × cos(angle)
Since the friction force and displacement are in opposite directions, the angle between them is 180°.
Therefore, the work done by the friction force is:
work = (μ × normal force) × distance × cos(180°)
work = (0.2 × 49 N) × 5 m × cos(180°)
Now, if we substitute the values into the equation:
work = (0.2 × 49 N) × 5 m × cos(180°)
work = 49 N × 5 m × (-1)
So, the work done by the friction force is:
work = -245 joules
To find the total work done on the box, we need to sum up the work done by the applied force and the work done by the friction force:
total work = work done by applied force + work done by friction force
total work = 50 joules + (-245 joules) = -195 joules
So, the total work done on the box is -195 joules.