# Calculate when a dock worker applies a constant horizontal force of 80.0 Newton to a block of ice on a smooth horizontal floor. If the frictional force is negligible, the block starts from rest and moves 11.0 meters in 5 seconds (a) What is the mass of the block of ice?(b) If the worker stops pushing at the end of 5 seconds, how far does the block move in the next 5 seconds?

(a)

Newton's 2nd Law

F = ma

Since there's no opposing force on the block of ice, the net force on the block is F = 80N

So 80 = ma or m = 80/a

To find m, we need to find a

Using Newton's equations of motion:

Initial velocity u = 0

Distance s = 11m

Time t = 5 seconds

Use s = ut + 1/2 at² because it's the only equation which gives us the acceleration a, while knowing all the other variables.

Substituting gives:

11 = (0)(5) + 1/2a(5²)

Rearranging:

11 = (1/2)a(25)

So:

a = 22/25 m/s²

Substituting in the equation m = 80/a gives:

m = 80 / (22/25) or m = 90.9 kg approx

(b)

Since there's no further acceleration (the worker stops pushing), and there's no deceleration (friction is negligible), the block will move at constant velocity (Newton's first law of motion).

So:

Use s = ut + 1/2 at² again

Since a = 0

s = ut + 1/2 (0)t²

or

s = ut

But we don't know the initial velocity u that the block travels at after the worker stops pushing. So first we have to go back and find it using the first equation of motion. We need to find v the final velocity after pushing and this will become the initial velocity u after pushing stops:

v = u + at

Substituting gives:

v = 0 + at = 0 + (22/25)5 = 110/25 = 22/5 m/s

So after the worker stops pushing

V = 22/5 m/s so u = 22/5 m/s

t = 5 s

a = 0 m/s²

Now substitute into s = ut + 1/2 at²

s = (22/5)(5) + (1/2)(0)(5² )

Or s = 22 m

Updated on March 10, 2018

### Original Article:

By Eugene Brennan

working