A dock worker applies a constant horizontal force of 80.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. the block starts from rest and moves 11.0 m in the first 5.00s what is the mass of the block of ice?
Calculate a from
S=ut + at2/2
where
S=11 m
u=0
t=5 s
From F=ma
where
F=80 N
a= as calculated above
you would be able to find m.
F = m a
a = 80/m
v = a t = (80/m) t
d = (1/2) a t^2 = (1/2)(80/m) t^2
11 = (1/2) (80/m) (25)
22 m = 80 * 25
I guess you can take it from there.
ljdc
To find the mass of the block of ice, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). In this case, we need to find the mass.
We are given the following information:
Force (F) = 80.0 N
Distance (d) = 11.0 m
Time (t) = 5.00 s
First, let's calculate the acceleration of the block using the formula:
a = (final velocity - initial velocity) / time
Since the block starts from rest, the initial velocity is zero. The final velocity can be found using the formula:
v = d / t
Substituting the given values:
v = 11.0 m / 5.00 s = 2.20 m/s
Now we can calculate the acceleration:
a = (2.20 m/s - 0) / 5.00 s = 0.44 m/s^2
Next, we can use the formula Newton's second law formula:
F = m * a
Rearranging the formula, we can solve for mass (m):
m = F / a
Substituting the given values:
m = 80.0 N / 0.44 m/s^2 ≈ 181.82 kg
Therefore, the mass of the block of ice is approximately 181.82 kg.