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You are pushing a 20-kg box along a horizontal floor. Friction acts on the box. When you apply a horizontal force of magnitude 48 N, the box moves at a constant velocity. If you increase your horizontal pushing force to 100 N, what is the acceleration of the box?

Answer :

xero099

Answer:

[tex]a = 2.6\,\frac{m}{s^{2}}[/tex]

Explanation:

The first box has the following equation of equilibrium:

[tex]\Sigma F = F - f = 0[/tex]

[tex]f = F[/tex]

[tex]f = 48\,N[/tex]

The coefficient of friction is:

[tex]\mu_{k} = \frac{f}{m \cdot g}[/tex]

[tex]\mu_{k} = \frac{48\,N}{(20\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]

[tex]\mu_{k} = 0.245[/tex]

There is a net acceleration, when horizontal pushing force is increased:

[tex]\Sigma F = F - f = m\cdot a[/tex]

[tex]a = \frac{F-f}{m}[/tex]

[tex]a = \frac{100\,N-48\,N}{20\,kg}[/tex]

[tex]a = 2.6\,\frac{m}{s^{2}}[/tex]

Explanation:

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