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Sagot :
Answer:
3.038 m/s^2
Explanation:
What we have
m = 6.0 kg
[tex]\mu_{k}[/tex] = 0.20
[tex]\mu_{s}[/tex] = 0.40
[tex]F_{applied}[/tex] = 30 N
Since the situation is that the box has already been pushed, lets get the total net force of both components.
∑Fy = [tex]ma_{y}[/tex] = 0 (bc the normal and weight cancel out each other)
∑Fx = [tex]F_{applied}[/tex] - [tex]f_{k}[/tex] = [tex]ma_{x}[/tex] (We are going to use this equation to find acceleration)
To find the friction force
[tex]f_{k}[/tex] = [tex]\mu_{k}[/tex] * N
[tex]f_{k}[/tex] = .20 * 6.0 kg * 9.81 m/s^2 = 11.772 N
Now we can use ∑Fx equation to actually find the acceleration of the box!
∑Fx = [tex]F_{applied}[/tex] - [tex]f_{k}[/tex] = [tex]ma_{x}[/tex]
Plug it in
30 N - 11.772 N = 6 kg * a
Do the algebra...
a = [tex]\frac{30 N - 11.772 N}{6 kg}[/tex] = 3.038 m/s^2
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