A dockworker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time 4.50 s. What is the mass of the block of ice?If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.70 s?
Use the statement
" The block starts from rest and moves a distance 13.0 m in a time 4.50 s. "
to solve for the block's acceleration, a.
X = (a/2)t^2
a = 2X/t^2
Using a, F = 79 N and F = m a, solve for m
The velocity when pushing stops at t = 4.5 s is
Vmax = a t
After 4.7 seconds, the additional distance traveled is Vmax*4.7 s
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To find the mass of the block of ice, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
We are given that the dockworker applies a constant horizontal force of 79.0 N to the block of ice. Since there is no frictional force, this force is equal to the net force acting on the block of ice.
We also have the distance moved by the block of ice, which is 13.0 m and the time taken, which is 4.50 s.
First, we can find the acceleration of the block of ice using the equation:
acceleration = distance / time
acceleration = 13.0 m / 4.50 s
acceleration = 2.89 m/s^2
Now, using Newton's second law of motion, we can find the mass of the block of ice:
force = mass x acceleration
79.0 N = mass x 2.89 m/s^2
mass = 79.0 N / 2.89 m/s^2
mass = 27.33 kg
Therefore, the mass of the block of ice is approximately 27.33 kg.
To find how far the block moves in the next 4.70 s after the worker stops pushing, we need to find the block's initial velocity and then calculate the distance using the equation:
distance = initial velocity x time + 0.5 x acceleration x time^2
Since the block starts from rest, the initial velocity is zero.
The acceleration of the block is already known, and the time is given as 4.70 s.
Plugging in these values, we have:
distance = 0 x 4.70 s + 0.5 x 2.89 m/s^2 x (4.70 s)^2
distance = 0 + 0.5 x 2.89 m/s^2 x 22.09 s^2
distance = 32.05 m
Therefore, the block moves approximately 32.05 m in the next 4.70 s after the worker stops pushing.
To find the mass of the block of ice, we can use Newton's second law of motion. The equation is given by F = ma, where F is the applied force, m is the mass, and a is the acceleration.
In this case, the applied force is the horizontal force exerted by the dockworker, which is 79.0 N. The block of ice starts from rest, so its initial velocity is 0 m/s. The block moves a distance of 13.0 m in a time of 4.50 s. We can use the equation of motion s = ut + 0.5at^2 to find the acceleration, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Substituting the given values, we have 13.0 = 0 + 0.5a(4.50^2). Solving this equation, we find that the acceleration is approximately 1.021 m/s^2.
Now, we can use Newton's second law to find the mass. Rearranging the equation F = ma, we get m = F / a. Substituting the values, we have m = 79.0 N / 1.021 m/s^2 = 77.39 kg.
Therefore, the mass of the block of ice is approximately 77.39 kg.
To find how far the block moves in the next 4.70 s after the worker stops pushing, we can use the equation of motion s = ut + 0.5at^2 again. This time, the initial velocity is the final velocity at the end of the previous 4.50 s (which we can calculate using the acceleration), and the time is 4.70 s.
First, let's find the final velocity vf using the equation vf = u + at. Since the initial velocity u is 0 m/s, we have vf = 0 + 1.021 m/s^2 * 4.50 s = 4.5955 m/s.
Now, we can use the equation of motion s = ut + 0.5at^2 with vf as the initial velocity and 4.70 s as the time. Substituting the values, we have s = 4.5955 m/s * 4.70 s + 0.5 * 1.021 m/s^2 * (4.70 s)^2.
Solving this equation, we get s = 10.176 m.
Therefore, the block moves approximately 10.176 m in the next 4.70 s after the worker stops pushing.