what is the acceleration produced by the resultant force acting on an object if the coefficient of friction acting between the b

Question

3 years ago

11

lock and surface is 0.1

1 answer:

LenKa [72] · Answer 1 · 2022-03-05T14:30:46+03:00

The acceleration of the block is $-0.98 m/s^2$

Explanation:

The expression for the force of friction acting on the block is (assuming the surface is horizontal and flat):

$F_f = -\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the block

g is the acceleration of gravity

and where the negative sign means the direction of the force is opposite to that of the motion of the block

If this is the only force acting on the object, then this is also the resultant force, so we can rewrite Newton's second law as:

$F=ma\\-\mu mg = ma$

where

a is the acceleration of the block

Re-arranging the equation,

$a=-\mu g$

And so this is the expression for the acceleration of the block acted upon the force of friction.

In this problem, we have:

$\mu=0.1$

$g=9.8 m/s^2$

Solving,

$a=-(0.1)(9.8)=-0.98 m/s^2$

Learn more about friction:

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