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.
a. What is the acceleration of the box while the man is pushing it?
b. What is the position of the box at the end of an interval of 3.00s? (assume x₀ = 0)
c. 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?
376 - 214 = F
F = m a
so
a = F/90.8
t = 3 s
v = a t = 3 a
x = 0 + 0 + (1/2) a t^2 = 4.5 a
To answer these questions, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
a. To find the acceleration of the box while the man is pushing it, we need to calculate the net force acting on the box. The net force is the force applied by the man minus the friction force.
Net force = Force applied by the man - Friction force
= 376N - 214N
= 162N
Using Newton's second law, we can calculate the acceleration:
Acceleration = Net force / Mass
= 162N / 90.8kg
≈ 1.78 m/s^2
Therefore, the acceleration of the box while the man is pushing it is approximately 1.78 m/s^2.
b. To find the position of the box at the end of a 3.00s interval, we can use the equations of motion. Since the box starts at rest, we can use the equation:
Position = Initial position + Initial velocity * Time + (1/2) * Acceleration * Time^2
The initial position (x₀) is given as 0 since we assume the box is at rest initially. The initial velocity can be calculated using the equation:
Initial velocity = Acceleration * Time
Substituting the given values into the equations:
Initial velocity = 1.78 m/s^2 * 3.00s
≈ 5.34 m/s
Position = 0 + 5.34 m/s * 3.00s + (1/2) * 1.78 m/s^2 * (3.00s)^2
Calculating this:
Position ≈ 48.15 m
Therefore, the position of the box at the end of a 3.00s interval is approximately 48.15 m.
c. If the man stops pushing the box at the end of the 3.00s time interval, the only force acting on the box is the friction force. Since the box comes to a stop, we know the acceleration is 0.
Using Newton's second law, we can find the friction force needed to bring the box to a stop:
Net force = Mass * Acceleration
0N = 90.8kg * Acceleration
Solving for acceleration:
Acceleration = 0N / 90.8kg
Therefore, the acceleration of the box as it comes to a stop is 0 m/s^2.