A box is given a push so that it slides across the floor. How far will it go, given the coefficient of kinetic friction is .21 and the push imparts an inital speed of 4.2 m/s?
Frictional force opposing movement
=μmg
Acceleration, a
=-μmg/m
=μg
initial velocity, v0 = 4.2 m/s
final velocity v1=0
v²-v0⊃ = 2aS
S=distance moved before box stops.
2.2
Francesca who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from JFK Airport. She notices that the string makes an angle of 25 degrees with respect to the vertical as the aircraft accelerates for takeoff , which takes about 18s. Estimate the takeoff speed of the aircraft.
To determine how far the box will go, we need to use the equations of motion along with the concept of kinetic friction.
The equation we need is:
\(d = \frac{{v_0^2}}{{2 \mu g}}\)
Where:
- \(d\) is the distance the box will travel
- \(v_0\) is the initial speed of the box
- \(\mu\) is the coefficient of kinetic friction
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²)
Substituting the given values into the equation:
\(d = \frac{{(4.2 \, \text{m/s})^2}}{{2 \times 0.21 \times 9.8 \, \text{m/s}^2}}\)
Calculating this expression will give us the distance the box will go.