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 0.16 and the push imparts an initial speed of 4.2 m/s?
initial KE= work done on friction
1/2 mv^2=mu*mg*distance
solve for how far.
To determine how far the box will go, we need to consider the force of kinetic friction acting on the box.
The force of kinetic friction can be calculated using the formula:
Fk = μk * N
Where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
The normal force is the force exerted by the floor on the box perpendicular to the surface. For a box on a flat floor, the normal force is equal to the weight of the box, given by:
N = m * g
Where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s²).
However, since the box is given an initial speed of 4.2 m/s, it experiences an opposing force due to the initial push. This force is equal to the product of the mass and acceleration, given by:
Fpush = m * a
To calculate the acceleration, we can use the following equation derived from Newton's second law of motion:
Fpush - Fk = m * a
Rearranging this equation, we get:
a = (Fpush - Fk) / m
Now we can substitute the values given in the problem:
μk = 0.16
initial speed (v0) = 4.2 m/s
m = ? (not given in the problem)
We also know that the initial speed is equal to the acceleration multiplied by the time taken (Δt), so we can use this relationship to find the time:
v0 = a * Δt
Solving for Δt:
Δt = v0 / a
By substituting the given values of v0 and a, we can calculate Δt.
Once we have the value of Δt, we can calculate the distance using the equation:
d = v0 * Δt
Substituting the values of v0 and Δt, we can determine the distance the box will go.