A dockworker applies a constant horizontal force of 73.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 11.5m in a time 5.40s .
a.) What is the mass of the block?
b.)If the worker stops pushing at the end of 5.40s , how far does the block move in the next 4.20s ?
a. d = Vo*t + 0.5a*t^2 = 11.5 m.
0 + 0.5a*5.4^2 = 11.5
14.58a = 11.5
a = 0.789 m/s^2
m = F/a = 73/0.789 = 92.55 kg.
b. V=Vo + a*t = 0 + 0.789*5.4=4.26 m/s.
d = Vo*t = 4.26 m/s * 4.20s = 17.89 m.
To solve this problem, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. We can also use the equation for displacement, which is given by the equation s = ut + 0.5at^2, where s is displacement, u is initial velocity, t is time, and a is acceleration.
Let's find the mass of the block first.
a) We know that a constant horizontal force of 73.0N is applied to the block and that the frictional force is negligible. Therefore, the net force on the block is 73.0N.
Using Newton's second law of motion, we can write:
Net force = Mass * Acceleration
73.0N = Mass * Acceleration
We know that the acceleration is given by the equation a = (final velocity - initial velocity) / time. In this case, the block starts from rest, so the initial velocity is 0 m/s. Given that the block moves a distance of 11.5m in a time of 5.40s, we can calculate the acceleration:
acceleration = (11.5m - 0m) / 5.40s
Now we can substitute this acceleration value into the equation for net force:
73.0N = Mass * [(11.5m - 0m) / 5.40s]
Simplifying the equation, we can rearrange it to solve for the mass:
Mass = (73.0N * 5.40s) / 11.5m
Solving this expression will give us the mass of the block.
Now let's move on to part b.
b) In this case, the worker stops pushing at the end of 5.40s, but the block will continue to move due to its inertia. We need to find out how far the block will move in the next 4.20s.
To solve this problem, we can still use the equation for displacement, s = ut + 0.5at^2. The initial velocity is the final velocity from the previous scenario (after 5.40s), and the time is 4.20s.
To find the final velocity after 5.40s, we can use the equation v = u + at, where u is the initial velocity and a is the acceleration calculated earlier.
The initial velocity is 0 m/s, and the acceleration is the same as before.
Now we can calculate the final velocity and use it in the equation for displacement:
s = (final velocity from previous scenario) * 4.20s + 0.5 * (acceleration from previous scenario) * (4.20s)^2
Solving this equation will give us the distance the block moves in the next 4.20s.