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a 3.5 kg block is pushed along a horizontal floor by a force : f of magnitude 15 n at an angle ???? 40° with the horizontal (fig. 6-19). the coefficient of ki- netic friction between the block and the floor is 0.25. calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.

Sagot :

The frictional force is 11 N and the block’s acceleration is 0.14 m/s².

What is the coefficient of friction?

It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.

For part (a):

Taking x component:

F(x) = F cosθ

Taking y component:

F(y) = -F sinθ

Applying Newton’s second law to the y-axis:

F(N) = 15 sin40 + (3.5)(9.8)

F(N) = 44 N

coefficient U = 0.25

f(k) = 11 N

For part (b):

Applying Newton’s second law to the x-axis:

a = ((15) cos40 - 11)/3.5

a = 0.14 m/s²

Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².

Learn more about the coefficient of friction here:

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