A 200g box is moving on a table with velocity of 23m/s. There is a coefficient of friction of 0.032 between the table an the box. A force pushes the box for 1.2m which increases the velocity to 28m/s. Calculate the force that was applied.
How do we solve this using energies and work? Not using Kinematics.
To solve this problem using energies and work, we can make use of the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.
The work done can be calculated using the equation:
Work = Force * Distance
Let's assume that the force applied is F.
Initially, the box is moving with a velocity of 23 m/s. The initial kinetic energy (KEi) of the box can be calculated using the equation:
KEi = (1/2) * m * v^2
where m is the mass of the box (200g = 0.2kg) and v is the initial velocity (23 m/s).
KEi = (1/2) * 0.2kg * (23 m/s)^2
= 115 J
Next, the force is applied to the box over a distance of 1.2 m, and the velocity increases to 28 m/s. The final kinetic energy (KEf) of the box can be calculated using:
KEf = (1/2) * m * v^2
where v is the final velocity (28 m/s).
KEf = (1/2) * 0.2kg * (28 m/s)^2
= 156.8 J
The work done on the box is equal to the change in kinetic energy:
Work = KEf - KEi
= 156.8 J - 115 J
= 41.8 J
Therefore, the work done on the box is equal to 41.8 J.
Now, we can find the force applied using the equation:
Work = Force * Distance
Rearranging the equation, we get:
Force = Work / Distance
Substituting the values, we have:
Force = 41.8 J / 1.2 m
= 34.83 N
Hence, the force applied to the box is approximately 34.83 N.