A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.30 and the push imparts an initial speed of 4.0 <units>m/s</units>?

Initial KE=mu*mg*distance

solve for distance. Notice mass m divides out.

To determine how far the box will go, we need to take into account the force of friction acting on the box.

First, let's identify the forces at play. When the box is pushed, it experiences an initial force in the direction of motion. However, as it starts to move, the force of kinetic friction opposes its motion.

The force of kinetic friction can be calculated using the equation:

\( f_{\text{friction}} = \mu_k \times N \)

where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.

The normal force (\( N \)) is the force exerted by the surface on the box perpendicular to the surface. In this case, since the box is sliding on the floor, the normal force will be equal to the weight of the box, which can be calculated using:

\( N = m \times g \)

where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity.

Now, the force of kinetic friction will act in the opposite direction to the initial force and cause deceleration of the box. The magnitude of the force of kinetic friction is given by:

\( f_{\text{friction}} = m \times a \)

where \( a \) is the deceleration.

Since we know the force of friction, we can calculate the deceleration (\( a \)) using:

\( a = \frac{{ f_{\text{friction}} }}{m} \)

Now, the distance (\( d \)) that the box will travel can be determined using the equation:

\( d = \frac{{ v_0^2 }}{{ 2a }} \)

where \( v_0 \) is the initial speed of the box.

Let's plug in the given values to calculate the distance traveled by the box.

Given:
\( \mu_k = 0.30 \)
\( v_0 = 4.0 \, \text{m/s} \)

Assume:
\( m = 1 \, \text{kg} \) (arbitrary mass for calculation purposes)

First, calculate the normal force:
\( N = m \times g \)
\( N = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
\( N = 9.8 \, \text{N} \)

Next, calculate the force of friction:
\( f_{\text{friction}} = \mu_k \times N \)
\( f_{\text{friction}} = 0.30 \times 9.8 \, \text{N} \)
\( f_{\text{friction}} = 2.94 \, \text{N} \)

Then, calculate the deceleration:
\( a = \frac{{ f_{\text{friction}} }}{m} \)
\( a = \frac{{ 2.94 \, \text{N} }}{1 \, \text{kg} } \)
\( a = 2.94 \, \text{m/s}^2 \)

Finally, calculate the distance traveled:
\( d = \frac{{ v_0^2 }}{{ 2a }} \)
\( d = \frac{{ 4.0^2 }}{{ 2 \times 2.94 }} \)
\( d \approx 2.74 \, \text{m} \)

Therefore, the box will travel approximately 2.74 meters.