physics help pleaseeee.?

Question

A 20 kg box was given a push and is sliding along a table with a coefficient of friction of 0.4.



What is the net force acting on the box?

What is the acceleration of the box?


please help!! thank you .

Answer 1

READ THIS CAREFULLY
As the block just slides on the table the net force just equals the limiting friction acting on the body

hence,
F=coeff. of friction * Nor,mal rxn

normal rxn= weight of body on a plane

hence F= 20*9.8*0.4
=78.4N

ACC
it is just begg to slide and the given data is just unsufficient 2 find the acc. of the box

Answer 2

A push is the applied force.

Frictional force acts opposite to the applied force.

Net force is the resultant of these two forces.

When the net force is zero, the box just begins to slide.

And when the net force is zero and if the box begins sliding it will move with constant speed and hence the acceleration will be zero.

Only when the applied force is greater than the frictional force the box will have acceleration.

The acceleration is found by dividing the excess force over the frictional force by the mass of the box.

In the given problem the frictional force is 0.4 x 20 x 9.8 = 78.4 N.

Since the body slides with constant speed the applied force is also 78.4 N.

If the applied force is more than 78.4, then the box will have acceleration.

Answer 3

f = mu*N
f = 0.4*(20*9.8)
f = 78.4 newtons

F = ma
78.4 = 20 a
a = 3.92 m/sec^2 deceleration or - 3.92 m/sec^2 acceleration

Answer 4

go with andy