Question

A `15` `kg` block of steel is at rest on a smooth, horizontal, icy surface. What net force must applied to the block so that it accelerates at `0.6``m``/``s^2`?

Answer 1

`F_{n et} =9` `N`

Explanation:

The question asks for the required net force for a particular acceleration. The equation that relates the net force to the acceleration is Newton's 2nd Law,

`F_{n et} =m a`,

where `F_{n et}` is the net force normally in Newtons, `N`;

`m` is the mass, in kilograms, `kg`;

and `a` is the acceleration in meters per second squared, `m/s^2`.

We have `m=15` `kg` and `a=0.6` `m/s^2`, so

`F_{n et} =(15` `kg )* (0.6` `m/s^2` `)= (15*0.6) * (kg*m/s^2 )`

remember `1` `N=kg*m/s^2`

`F_{n et} =9` `N`