Answer:
The magnitude of the friction force exerted on the box is 2.614 newtons.
Explanation:
Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:
[tex]\Sigma F = -f = m\cdot a[/tex] (Eq. 1)
Where:
[tex]f[/tex] - Kinetic friction force, measured in newtons.
[tex]m[/tex] - Mass of the box, measured in kilograms.
[tex]a[/tex] - Acceleration experimented by the box, measured in meters per square second.
By applying definitions of weight ([tex]W = m\cdot g[/tex]) and uniform accelerated motion ([tex]v = v_{o}+a\cdot t[/tex]), we expand the previous expression:
[tex]-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex]
And the magnitude of the friction force exerted on the box is calculated by this formula:
[tex]f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex] (Eq. 1b)
Where:
[tex]W[/tex] - Weight, measured in newtons.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]W = 52.4\,N[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 1.37\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]t = 2.8\,s[/tex], the magnitud of the kinetic friction force exerted on the box is:
[tex]f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)[/tex]
[tex]f = 2.614\,N[/tex]
The magnitude of the friction force exerted on the box is 2.614 newtons.
