A wooden block of mass 3.270 kg sits on a metal table. The coefficients of stati

Question

A wooden block of mass 3.270 kg sits on a metal table. The coefficients of static and kinetic friction between the block and the table are us- 0.605 and pk 0.305. At time t- 0, a horizontal force F 11.9 N is exerted on the block. Give the force of friction exerted on the block by the table at the following times Hints Under what conditions does static friction apply? How about kinetic friction? Number Number 11.9 at is the initial speed of the block? Will the block accelerate? Incorrect. Perhaps you calculated the net force rather than the frictional force Consider the same situation, but this time the external force F is 24.1 N. Again state the force of friction acting on the block at the following times: Number Number 9.78

Explanation / Answer

Mass of wooden block = 3.27 kg

Normal force on block by table; N = m*g = 3.27 * 9.8 = 32.046 Newton

As horizontal force F = 11.9 N is exerted,

we can have maximum available static friction force = s * N ; if it is more than F

s * N = 0.605 * 32.046 = 19.39 N which is more than F = 11.9 N,

so, maximum static Friction force will be equal to F; Fs = 11.9 N for t = 0

Fs = 11.9 N for t > 0 as block will not move even after t>0

For part 2, F = 24.1 N

So, maximum static friction force = s * N = 0.605 * 32.046 = 19.39 N which is lower than F = 24.1 N

So, Fs = 19.39 N for t=0

As F > Fs; block will start moving and then kinteic friction force will be applied on block which can be calculated as below

Fk = k * N = 0.305 * 32.046 = 9.77 N for t>0

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