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 10.0m in the first 6.00s. What is the mass of the block of ice?
d = (1/2) a t^2
10 = (1/2)a (36)
a = 20/36
then
F = m a
80 = m (20/36)
so
m = 36 * 80 / 20
To find the mass of the block of ice, we can use Newton's second law of motion. According to the second law, the net force applied to an object is equal to the mass of the object multiplied by its acceleration.
In this case, we are given the applied force (80.0N) and the distance traveled (10.0m) in a given time interval (6.00s). We need to find the mass of the block of ice.
First, we can find the acceleration of the block using the kinematic equation:
distance = initial velocity × time + (1/2) × acceleration × time^2
Given that the block starts from rest, the initial velocity (u) is 0, and the equation simplifies to:
distance = (1/2) × acceleration × time^2
Plugging in the values, we have:
10.0m = (1/2) × acceleration × (6.00s)^2
Simplifying the equation, we find:
acceleration = (2 × distance) / (time^2)
acceleration = (2 × 10.0m) / (6.00s)^2
Next, we can find the mass of the block using Newton's second law. Rearranging the formula, we have:
mass = net force / acceleration
In this case, the net force is the applied force of 80.0N. Plugging in the values, we get:
mass = 80.0N / acceleration
Now, substitute the value of acceleration that we found earlier:
mass = 80.0N / ((2 × 10.0m) / (6.00s)^2)
Simplifying the equation:
mass = (80.0N × (6.00s)^2) / (2 × 10.0m)
Finally, we calculate the mass:
mass = 1440.0 kg
Therefore, the mass of the block of ice is 1440.0 kg.