A box which has a mass of 90.8kg is at rest on a rough horizontal floor, A man pushes on the box with a horizontal force of 376N. As soon as the box begins to move a friction force of 214N acts on the box from the floor.
If the man stops pushing the box at the end of the 3.00s time interval, what is the acceleration of the box as the box comes to a stop?
acceleration is ... f / m = net force / mass = friction force / mass
To find the acceleration of the box as it comes to a stop, we can use Newton's second law of motion:
F_net = m * a
Where:
- F_net is the net force acting on the box
- m is the mass of the box
- a is the acceleration of the box
In this scenario, the net force acting on the box is the difference between the force applied by the man and the friction force:
F_net = F_applied - F_friction
Using the given values, we have:
F_applied = 376N (force applied by the man)
F_friction = 214N (friction force)
Substituting these values into the equation, we get:
F_net = 376N - 214N
F_net = 162N
Now, let's substitute this value into Newton's second law and solve for acceleration:
162N = 90.8kg * a
To find the acceleration, divide both sides of the equation by the mass:
a = 162N / 90.8kg
a ≈ 1.78 m/s^2
Therefore, the acceleration of the box as it comes to a stop is approximately 1.78 m/s^2.