a dockworker applies a constant horizontal force of 80 N to a block of ice on a smooth horizontal floor. The friction force is negligible the block start from rest and move 11m in 5 sec what is it mass & if worker stop pushing at 5sec .How for does the block move in 5 ec

d = (1/2) a t^2

11 = (1/2) a (25)

a = .88 m/s^2

so
80 = m (.88)
m = 90.9 kg

v = a t = .88*5 = 4.4 m/s which continues without force
for five seconds so
4.4 * 5 = 22 meters more

Well, isn't it nice to hear that the friction force is being neglected? It's like the block of ice is on vacation from dealing with any resistance!

Now, let's get down to the calculations. We know that the dockworker applies a constant horizontal force of 80N to the block of ice. Since there is no friction, this force is enough to get the block moving.

Using Newton's second law (F = ma), we can determine the mass of the block. The force applied by the worker (80N) is equal to the mass of the block (m) multiplied by its acceleration (a).

So, 80N = m * a

Now, we know that the block starts from rest and moves a distance of 11m in 5 seconds. This allows us to calculate the acceleration of the block. The formula for distance covered can be rearranged as:

Distance (d) = Initial velocity (0) * time (t) + (1/2) * acceleration (a) * time (t)²

11m = 0 * 5s + (1/2) * a * (5s)²

11m = (1/2) * 25s² * a

22m = 25s² * a

a = 22m / (25s²)

Now that we have the acceleration, we can substitute it back into the initial equation to find the mass of the block:

80N = m * (22m / 25s²)

Simplifying this equation, we find that the mass of the block is approximately:

m ≈ 91.43 kg

Now, as for the second part of your question, if the worker stops pushing the block at 5 seconds, it will continue to move with the same velocity due to inertia. Therefore, the block will move the same distance as it did in the previous 5 seconds, which is 11 meters. So, it'll have traveled a total distance of 11 meters in 5 seconds.

I hope I was able to put a smile on your face with my calculations!

To find the mass of the block of ice, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, we need to calculate the acceleration of the block. We know that the block starts from rest and moves a distance of 11 meters in 5 seconds. We can use the formula:

s = ut + (1/2)at^2

Where:
s = displacement (11 meters)
u = initial velocity (0 m/s)
t = time (5 seconds)
a = acceleration

Plugging in the given values, we can solve for acceleration:

11 = 0 + (1/2)a(5^2)
11 = (1/2)25a
11 = 12.5a
a = 11 / 12.5
a ≈ 0.88 m/s^2

Now that we have the acceleration, we can use Newton's second law:

F = ma

Given that the dockworker applies a constant horizontal force of 80 N, we can substitute the values into the equation:

80 = m * 0.88

Solving for the mass:

m = 80 / 0.88
m ≈ 90.91 kg

Therefore, the mass of the block of ice is approximately 90.91 kg.

Next, if the worker stops pushing at 5 seconds, the block will continue moving due to its inertia. Considering there is no friction, the block will continue moving at a constant velocity.

Using the formula:

s = ut + (1/2)at^2

Where:
s = displacement (unknown)
u = initial velocity (0 m/s)
t = time (5 seconds)
a = acceleration (0.88 m/s^2)

Since we are looking for the displacement, we can rearrange the formula:

s = ut + (1/2)at^2
s = 0(5) + (1/2)(0.88)(5)^2
s = (1/2)(0.88)(25)
s = 11 meters

Therefore, the block will move an additional 11 meters in the next 5 seconds, even after the worker stops pushing.

To find the mass of the block, 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.

First, we need to calculate the acceleration of the block using the given information. We can use the equation:

acceleration = change in velocity / time

The block starts from rest, so its initial velocity is 0 m/s. The final velocity can be calculated using the equation:

final velocity = initial velocity + (acceleration * time)

Given that the block moves 11 m in 5 sec, we can calculate:

final velocity = 11 m / 5 s = 2.2 m/s

Since the block starts from rest, the initial velocity is 0 m/s. Therefore, the change in velocity is:

change in velocity = final velocity - initial velocity
= 2.2 m/s - 0 m/s
= 2.2 m/s

Using the equation for acceleration:

acceleration = change in velocity / time
= 2.2 m/s / 5 s
= 0.44 m/s^2

Now, we can use Newton's second law to find the mass of the block:

force = mass * acceleration
80 N = mass * 0.44 m/s^2

Solving for mass:

mass = 80 N / 0.44 m/s^2
mass ≈ 181.82 kg

Therefore, the mass of the block is approximately 181.82 kg.

Now, if the worker stops pushing at 5 seconds, the block would continue to move due to its inertia. Since no friction force is acting on the block, it will move with a constant velocity. We can find the distance traveled by multiplying the constant velocity by the time.

Since the final velocity is the same as the velocity calculated before (2.2 m/s), and the time is 5 seconds, we can calculate:

distance = velocity × time
distance = 2.2 m/s × 5 s
distance = 11 m

Therefore, if the worker stops pushing at 5 seconds, the block would move an additional 11 meters during that time.

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