A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of kinetic friction is 0.30.
a. What is the minimum horizontal force needed to get the box moving?
b. If you continue to push with that force, what will the acceleration of the box be?
Please help asap. Thank you.
a. .45*mg
b. .45*mg-.30mg=ma
a= .15g
To find the minimum horizontal force needed to get the box moving, we can use the equation for static friction:
\(F_{\text{static}} = \text{coefficient of static friction} \times \text{normal force}\)
where \(F_{\text{static}}\) is the maximum force of static friction that can be overcome before the box starts moving.
The normal force, \(N\), is equal to the weight of the box, \(mg\), where \(m\) is the mass of the box and \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
a. Calculating the maximum force of static friction:
\(F_{\text{static}} = \text{coefficient of static friction} \times N\)
\(F_{\text{static}} = 0.45 \times (500 \text{ kg} \times 9.8 \text{ m/s²})\)
\(F_{\text{static}} = 0.45 \times (4900 \text{ N})\)
\(F_{\text{static}} = 2205 \text{ N}\)
Therefore, the minimum horizontal force needed to get the box moving is 2205 N.
b. If you continue to push with that force, the box will experience kinetic friction, which is given by:
\(F_{\text{kinetic}} = \text{coefficient of kinetic friction} \times N\)
\(F_{\text{kinetic}} = 0.30 \times (500 \text{ kg} \times 9.8 \text{ m/s²})\)
\(F_{\text{kinetic}} = 0.30 \times (4900 \text{ N})\)
\(F_{\text{kinetic}} = 1470 \text{ N}\)
The net force acting on the box is equal to the applied force minus the force of kinetic friction:
\(F_{\text{net}} = F_{\text{applied}} - F_{\text{kinetic}}\)
Since the applied force is equal to the force of kinetic friction, the net force acting on the box is zero. This means the box will not accelerate and will move at a constant velocity when pushed with a force of 2205 N.