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A 12-kg bag of groceries is tossed onto a table at 4.0 m/s and slides to a stop in 2.2 s .
Modify the equation FΔt=Δ (mv) to find the force of friction.

Respuesta :

As we know that bag is stopped due to the force of friction on the table

so here by the given equation we will say

[tex]F\Delta t = \Delta (mv)[/tex]

now we have

[tex]F = \frac{\Delta (mv)}{\Delta t}[/tex]

now from the above equation we also have

[tex]F = \frac{(mv)_f - (mv)_i}{\Delta t}[/tex]

now we have

[tex]v_f = 0[/tex]

[tex]v_i = 4 m/s[/tex]

m = 12 kg

[tex]\Delta t = 2.2 s[/tex]

now we will have

[tex]F = \frac{0 - (12)(4)}{2.2}[/tex]

[tex]F = -21.82 N[/tex]

so friction force will be 21.82 N

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