A dockworker applies a constant horizontal force of 71.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 10.0m in a time of 5.50s .
If the worker stops pushing after 5.50s , how far does the block move in the next 5.10s ?
d = 0.5a*t^2 = 10 m.
0.5a*(5.5)^2 = 10
15.125a = 10
a = 0.661 m/s^2.
V = a*t = 0.661 * 5.5 = 3.636 m/s.
d = Vo*t - o.5a*t^2
d=3.636*5.10 - 0.5*0.661*5.10^2=9.95 m.
To determine how far the block moves in the next 5.10s, we need to understand the concept of acceleration and how it relates to displacement.
First, let's calculate the acceleration of the block using the given information. We can use the kinematic equation:
Displacement = Initial Velocity × Time + (1/2) × Acceleration × Time^2
In this case, the initial velocity is 0 since the block starts from rest. We are given the displacement (10.0m) and the time (5.50s). Plugging in these values, we can solve for acceleration:
10.0m = 0 × 5.50s + (1/2) × Acceleration × (5.50s)^2
Simplifying the equation, we get:
5.0m = (1/2) × Acceleration × 5.50s^2
Now, solve for acceleration:
Acceleration = (2 × 5.0m) / (5.50s^2)
Acceleration ≈ 1.818m/s^2
Now that we know the acceleration, we can use it to find the block's velocity at the end of the 5.50s pushing period. Using another kinematic equation:
Final Velocity = Initial Velocity + Acceleration × Time
Since the block starts from rest, the initial velocity is 0. Plugging in the values, we can solve for the final velocity:
Final Velocity = 0 + 1.818m/s^2 × 5.50s
Final Velocity ≈ 10.0m/s
Now that we know the final velocity at the end of the 5.50s pushing period, we can find the displacement in the next 5.10s. Again, using the kinematic equation:
Displacement = Initial Velocity × Time + (1/2) × Acceleration × Time^2
This time, the initial velocity is the final velocity from the previous calculation (10.0m/s). We're given the time (5.10s). Plugging in these values, we can find the displacement:
Displacement = 10.0m/s × 5.10s + (1/2) × 1.818m/s^2 × (5.10s)^2
Simplifying the equation, we get:
Displacement ≈ 51.0m
Therefore, the block moves approximately 51.0m in the next 5.10s after the worker stops pushing.