Question

A 4kg table pushed to the right with an applied force of 50N. The table has a net acceleration of 10 m/s^2 to the right. What is the magnitude of the force of friction acting on the table?

Answer 1

Answer:

Force of friction is 10 N.

Explanation:

Given:

Mass of the table is, $m=4\textrm{ kg}$

Force acting towards right, $F=50\textrm{ N}$

Net acceleration of the table, $a=10\textrm{ }m/s^{2}$

Let the force of friction acting on the table to the left be $f$ N.

Therefore, net force acting on the table is given as,

$F_{net}=F-f=50-f$

Now, as per Newton's second law of motion,

$F_{net}=ma$

Equating the above two equations, we get

$50-f=ma\\f=50-ma\\f=50-(4\times 10)\\f=50-40=10\textrm{ N}$

Therefore, the magnitude of the force of friction acting on the table is 10 N to the left direction.