A box is pushed along a surface with constant horizontal force. The surface is frictionless. The box starts from rest and travels 6.0m in 2.0s. if the experiment is repeated but with the box's mass doubled, how far will it travel?
a = f / m ... d = 1/2 a t^2
the acceleration is inversely proportional to the mass
the distance is directly proportional to the acceleration
To solve this problem, we will apply Newton's second law of motion and the equations of motion for constant acceleration.
First, let's analyze the initial situation where the box travels 6.0m in 2.0s with a constant horizontal force.
We can start by finding the acceleration of the box. The equation connecting distance (d), initial velocity (u), time (t), and acceleration (a) is:
d = ut + (1/2)at^2
Since the box starts from rest, the initial velocity (u) is 0, and the equation simplifies to:
d = (1/2)at^2
Rearranging the equation gives us:
a = (2d) / t^2
Now we know that acceleration is equal to the applied force divided by the mass of the object. In this case, the surface is frictionless, so the only force acting on the box is the applied force. Therefore, the acceleration can also be expressed as:
a = F / m
Combining the equations for acceleration, we have:
F / m = (2d) / t^2
Simplifying the equation:
F = m(2d / t^2)
Now, let's move on to the second situation, where the box's mass is doubled. We can denote the new mass as 2m.
Comparing the equations for force between the two situations, we can see that the force will be doubled in the second case because the mass is doubled:
F2 = (2m)(2d / t^2) = 4m(2d / t^2) = 8d(m / t^2)
Now, we need to find the distance (d2) traveled by the box in the second situation. We can use the equation of motion:
d2 = ut2 + (1/2)a2t2^2
Since the box starts from rest, the initial velocity (u2) is 0:
d2 = (1/2)a2t2^2
The acceleration (a2) can be calculated using the equation:
a2 = F2 / (2m)
Substituting the value of F2 and simplifying the equation:
a2 = 8d(m / t^2) / (2m)
a2 = 4d / t^2
Plugging the value of a2 into the equation of motion:
d2 = (1/2)(4d / t^2)t2^2
d2 = 2d
Therefore, in the second situation where the box's mass is doubled, it will travel a distance of 2d, which is twice the distance traveled in the initial situation.
In summary:
- Initial situation: The box travels a distance of d.
- Second situation: If the box's mass is doubled, it will travel a distance of 2d.