How much power must you exert to horizontally drag a 21.0kg table 15.0m across a brick floor in 20.0s at constant velocity, assuming the coefficient of kinetic friction between the table and floor is 0.500?
See previous post: Thu,10-24-13,3:05 PM.
To calculate the power required to drag the table across the floor, we need to consider the work done and the time taken.
First, let's calculate the work done, which is the force exerted multiplied by the distance moved. The force required to drag the table is equal to the force of friction acting on it.
The friction force can be calculated using the equation:
Friction force = coefficient of friction * normal force,
where the normal force is the force perpendicular to the surface. In this case, since the table is on a horizontal floor, the normal force is equal to the weight of the table, which is given by:
Normal force = mass * gravity,
where the mass of the table is 21.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.
So, the friction force is:
Friction force = 0.500 * (21.0 kg * 9.8 m/s^2).
Next, we need to calculate the work done. The work done is given by the equation:
Work = force * distance.
In this case, the distance is given as 15.0 m.
Now, we can substitute the values into the equation:
Work = Friction force * distance = (0.500 * 21.0 kg * 9.8 m/s^2) * 15.0 m.
Finally, we need to calculate the power, which is the work done divided by the time taken:
Power = Work / time = (0.500 * 21.0 kg * 9.8 m/s^2 * 15.0 m) / 20.0 s.
Now, you can calculate the power required using the given values.