If John was pushing a 300 kg desk across the floor with a constant acceleration of 3.5 m/s squared, how much force would he have to apply to the desk to maintain the acceleration, considering the coefficient of kinetic friction is 0.60?
kokok
To calculate the force John would have to apply to the desk to maintain the given acceleration, we need to consider the force of friction opposing his push. The force of friction can be calculated using the equation:
\(F_{friction} = \mu \times N\)
Where:
\(F_{friction}\) is the force of friction,
\(\mu\) is the coefficient of kinetic friction,
and \(N\) is the normal force.
The normal force is the force exerted by a surface that is perpendicular to the surface. In this case, the desk is on a horizontal surface, so the normal force is equal to the weight of the desk, which can be calculated using the equation:
\(N = m \times g\)
Where:
\(N\) is the normal force,
\(m\) is the mass of the desk, and
\(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, the mass of the desk is given as 300 kg. Therefore:
\(N = 300 \times 9.8 = 2940\) N
Now that we know the normal force, we can calculate the force of friction using the given coefficient of kinetic friction (\(\mu = 0.60\)):
\(F_{friction} = 0.60 \times 2940 = 1764\) N
The force John would have to apply to the desk to overcome the friction and maintain the given acceleration can be calculated using Newton's second law:
\(F_{applied} - F_{friction} = m \times a\)
Rearranging this equation, we find:
\(F_{applied} = m \times a + F_{friction}\)
Substituting the given values, we have:
\(F_{applied} = 300 \times 3.5 + 1764 = 2838\) N
Therefore, John would have to apply approximately 2838 N of force to the desk to maintain the given acceleration, considering the coefficient of kinetic friction is 0.60.