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?
F = m*g = 21kg * 9.8N/kg = 205.8 N.
Fk = u*mg = 0.5 * 205.8 = 102.9 N.=Force
of kinetic friction.
Fe-Fk = m*a
Fe-102.9 = m*0 = 0
Fe = 102.9 N. = Force exerted.
Pe = Fe * d/t = 102.9 * 15/20 = 77.18 N.
= Power exerted.
Correction:
Pe=102.9 * 15/20 = 77.18 J/s=77.18 Watts
= Power exerted.
To calculate the power required to horizontally drag the table, we need to consider the work done in overcoming the force of kinetic friction. The formula for power is:
Power = Work / Time
First, let's calculate the work done. The work done against friction is given by the formula:
Work = Force * Distance
The force of friction can be calculated using the equation:
Force of friction = μ * Normal force
The normal force is the force exerted by the table on the floor, which is equal to the weight of the table. The formula for weight is:
Weight = mass * gravity
Now, we can substitute the values into the formulas:
Weight = 21.0 kg * 9.8 m/s^2 = 205.8 N
Force of friction = 0.500 * 205.8 N = 102.9 N
Next, we calculate the work done:
Work = 102.9 N * 15.0 m = 1543.5 J
Finally, we can use the calculated work and the given time to find the power:
Power = 1543.5 J / 20.0 s = 77.2 W
Therefore, the power required to horizontally drag the table is 77.2 watts.