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

Force of friction is 10 N.

Step-by-step 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* 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.