Questions: A 405-N rightward force is used to drag a large box across the floor with a constant velocity of 0.678 m / s. The coefficient of friction between the box and the floor is 0.795. Determine the mass of the box.

Solution Steps

Identify the forces acting on the box. The box is moving with a constant velocity, which implies that the net force acting on it is zero. The forces involved are the applied force, the frictional force, and the gravitational force.

Since the box moves with constant velocity, the applied force equals the frictional force. Therefore, the frictional force \( f \) can be expressed as: \[ f = 405 \, \text{N} \]

The frictional force is also given by the formula: \[ f = \mu \cdot N \] where \( \mu = 0.795 \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, the normal force \( N \) is equal to the gravitational force on the box, which is \( N = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).

Substitute the expression for the normal force into the friction formula: \[ 405 = 0.795 \cdot m \cdot 9.8 \] Solve for \( m \): \[ m = \frac{405}{0.795 \cdot 9.8} \]

Final Answer