A dockworker applies a constant horizontal force of 90.0Nto a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.0min a time of 5.20s. If the worker stops pushing after 5.20s, how far does the block move in the next 5.30s

Distance traveled in the next 5.30 seconds is 24.454 meters.

Explanation:

Considering the block moves with a constant acceleration when the force is applied, the average speed would be half of the speed at the end of 5.20 seconds of acceleration.

This means:

Average Speed = Total distance / Time

Average Speed = 12 / 5.20 = 2.307 m/s

Speed at end of 5.20 seconds = 2 * Average Speed

Speed at end of 5.20 seconds = 2 * 2.307

Speed at end of 5.20 seconds = 4.614 m/s

Now that we have the speed at which the block is traveling, and there is no friction to reduce or change the speed, we can solve for the distance covered in the next 5.30 seconds in the following way:

Distance traveled in the next 5.30 seconds:

Distance = Speed * Time

Distance = 4.614 * 5.30

Distance = 24.454 meters

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more