A block with a mass of 126 kg is pulled with a horizontal fource of F_{applied} across a rough floor. The coefficent of friction between the floor and the block is 0.7. If the block is moving at a constant velocity, what is the magnitude of F_{applied}?
F_{applied} =
weight - 126 g
friction force = 126 * 9.81 * .7
= F if no acceleration
i tried that and answer is incorrect....
To find the magnitude of F_{applied}, we need to consider the forces acting on the block and set up an equation that represents the equilibrium of forces.
In this case, the forces acting on the block are the force of gravity (weight) and the force of friction.
1. Force of gravity (weight):
The force of gravity can be calculated using the formula:
Weight = mass * acceleration due to gravity (9.8 m/s^2)
Given that the mass of the block is 126 kg, we can calculate the weight:
Weight = 126 kg * 9.8 m/s^2
2. Force of friction:
The force of friction can be calculated using the formula:
Force of friction = coefficient of friction * normal force
The normal force is the perpendicular force exerted by a surface, which is equal to the weight of the object in this case since the block is on a horizontal surface. Therefore, the normal force is equal to the weight of the block.
Using the formula for the force of friction, we can calculate it:
Force of friction = coefficient of friction * weight
Given that the coefficient of friction is 0.7, we can calculate the force of friction:
Force of friction = 0.7 * (mass * acceleration due to gravity)
3. Equilibrium of forces:
Since the block is moving at a constant velocity, the net force acting on it must be zero. Therefore, the magnitude of F_{applied} must be equal to the force of friction.
F_{applied} = Force of friction
Substituting the calculated force of friction into the equation, we can determine the magnitude of F_{applied}.