# Finding acceleration when given coefficients

what will be the rate of acceleration be when a 10N force is applied to a 7.0kg block of wood resting on a table with a coefficient of friction equalling 0.45?

ok so i am totally confused, im thinking i have to use 2 formulas but im not exactly sure

i tried the formula:
Fn = Fg = mg
and did:
Fn = 10(7.0) = 70N
then i thought:
Fκ = μ Fn
so i did:
Fκ = (0.45)(70) = 31.5m/s(squared)

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ehild
Homework Helper
If the force of friction exceeds the applied force the object can not start to move, can it? It is static friction that acts on an object in rest. The static friction is less than or equal to μ Fn: Fs≤μ Fn. Only when the applied force exceeds the maximum value of the static friction, can the object accelerate in the direction of the applied force.

ehild

You have taken g = 10 N/kg..... that is OK, so the weight of your block is 70N which gives a friction force of 0.45 x 70 = 31.5N (this is not an acceleration).
The applied force on the object is 10N which is less than the friction force. This indicates that the object is decellerating (-acceleration)
The resultant force is 31.5 - 10 = 21.5N and therefore the acceleration/decelleration is
a = F/m = 21.5/7 = 3.04 ms^-2
Should really be written as -3.04 ms^-2

PS
The wording of the question is not great.....'resting on a table'..... suggests not moving but the numbers do not support that !!!!
Where did the question come from.

thanks guys! it was on one of the practice tests that my teacher wrote out for us, this helped alot! :)