A 33.0 kg box initially at rest is pushed 5.95 m along a rough, horizontal floor with a constant applied horizontal force of 145 N. If the coefficient of friction between box and floor is 0.300.
find:
the work done by the applied force
the increase in internal energy in the box-floor system due to friction
the work done by the normal force
the work done by the gravitational force
the change in kinetic energy of the box
the final speed of the box
To solve this problem, we will use several physics principles and formulas.
1. Work done by the applied force:
The work done by a force is given by the formula:
Work = Force × Distance × cos(theta)
In this case, the applied force is horizontal, so the angle (theta) between the force and the displacement is 0 degrees.
Therefore, the work done by the applied force is:
Work = Applied force × Distance × cos(0 degrees)
= Applied force × Distance
Plugging in the values:
Applied force = 145 N
Distance = 5.95 m
Work = 145 N × 5.95 m
Work = 862.75 J
So, the work done by the applied force is 862.75 Joules.
2. Increase in internal energy in the box-floor system due to friction:
The work done by friction can be calculated using the formula:
Work = Force of friction × Distance
The force of friction can be found using the equation:
Force of friction = Normal force × Coefficient of friction
To find the normal force, we can use the equation:
Normal force = Weight of the box = mass × gravity
Plugging in the values:
Mass of the box = 33.0 kg
Coefficient of friction = 0.300
Gravity = 9.8 m/s^2
Weight of the box = 33.0 kg × 9.8 m/s^2
Normal force = 323.4 N
Force of friction = 323.4 N × 0.300
Force of friction = 97.02 N
Now, we can calculate the work done by friction:
Work = Force of friction × Distance
= 97.02 N × 5.95 m
= 577.439 J
So, the increase in internal energy in the box-floor system due to friction is 577.439 Joules.
3. Work done by the normal force:
Since the normal force acts perpendicular to the displacement, the angle (theta) between them is 90 degrees. Therefore, the work done by the normal force is:
Work = Normal force × Distance × cos(90 degrees)
= Normal force × Distance × 0
= 0 J
So, the work done by the normal force is 0 Joules.
4. Work done by the gravitational force:
The gravitational force acts vertically, while the displacement is horizontal. Therefore, the angle (theta) between them is 90 degrees. Hence, the work done by the gravitational force is also:
Work = Weight of the box × Distance × cos(90 degrees)
= Weight of the box × Distance × 0
= 0 J
So, the work done by the gravitational force is 0 Joules.
5. Change in kinetic energy of the box:
The change in kinetic energy can be calculated using the formula:
Change in kinetic energy = Work done by the applied force + Work done by friction
= 862.75 J + 577.439 J
= 1440.189 J
So, the change in kinetic energy of the box is 1440.189 Joules.
6. Final speed of the box:
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Therefore, we can equate the work done by the applied force to the change in kinetic energy:
Work done by the applied force = Change in kinetic energy
Applied force × Distance = (1/2) × mass × (final velocity)^2
Plugging in the values:
Applied force = 145 N
Distance = 5.95 m
Mass = 33.0 kg
Change in kinetic energy = 1440.189 J
145 N × 5.95 m = (1/2) × 33.0 kg × (final velocity)^2
Final velocity = sqrt((145 N × 5.95 m) / ((1/2) × 33.0 kg))
Final velocity = 5.32 m/s
So, the final speed of the box is 5.32 m/s.