dockworker applies a constant horizontal force of 80.0N to a block of ice on a smooth horizontal floor. thhe block starts from rest and moves 11.0m in 5.00s. (a) what is the mass of the ice block? (b) if the work stops pushing at the end of the 5.00s, How far does the block move in the next 5.00s?

distance=1/2 a t^2=1/2 force/mass *t^2

do the algebra, solve for mass.

b. No friction? It moves forever at constant velocity. Maybe it ends up in Neverland, who knows.

To find the answers to these questions, we can use the concepts of Newton's second law and work-energy principle.

(a) To determine the mass of the ice block, we need to use Newton's second law, which states that force equals mass times acceleration (F = m*a). In this case, the force applied by the dockworker is 80.0N, and the acceleration can be determined using the equation of motion:

s = ut + (1/2)at^2

where
s = displacement (11.0m),
u = initial velocity (0 m/s),
t = time (5.00s).

Rearranging the equation gives us:

a = 2s / t^2

Substituting the values:

a = 2*(11.0m) / (5.00s)^2

Now we can use Newton's second law to find the mass:

m = F / a

Substituting the force value:

m = 80.0N / (2*(11.0m) / (5.00s)^2)

Simplifying the expression, we can calculate the mass.

(b) To determine how far the block moves in the next 5.00s, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The work done by the dockworker is given by:

work = force * distance

Since the work done equals the change in kinetic energy, we can write:

work = (1/2) * m * v^2

where
work = force * distance,
m = mass (which we calculated in the previous step),
v = final velocity.

We can rearrange this equation to find the final velocity:

v^2 = (2 * work) / m

Given that no additional work is done on the block after the 5.00s, the work done is equal to the force applied by the dockworker multiplied by the distance moved in the first 5.00s (80.0N * 11.0m). Substituting this value, along with the calculated mass, we can solve for v.

Once we have the final velocity, we can use the equation of motion s = ut + (1/2)at^2 to find the distance moved in the next 5.00s. In this case, the initial velocity is the final velocity at the end of the first 5.00s, and the time is again 5.00s.

With these steps, we can calculate the mass of the ice block and the distance it moves in the next 5.00s.