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?



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)


11 = (1/2)a(25)


a = 22/25 m/s²

Substituting in the equation m = 80/a gives:

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


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).


Use s = ut + 1/2 at² again

Since a = 0

s = ut + 1/2 (0)t²


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:

Force, Weight, Newtons, Velocity, Mass and Friction—Basic Principles of Mechanics
By Eugene Brennan