A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.4 s.
a.What is the magnitude of the average net force on the box?
b.You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s. How long does it take to stop?
Düsseldorf
a. To find the magnitude of the average net force on the box, we can use Newton's second law, which states that the net force is equal to the product of the mass and acceleration.
Given:
Mass of the box (m) = 6 kg
Initial speed (u) = 5 m/s
Final speed (v) = 0 m/s (the box comes to a stop)
Time (t) = 1.4 s
Using the equation of motion:
v = u + at
Rearranging the equation:
a = (v - u) / t
Substituting the given values:
a = (0 - 5) m/s / (1.4 s) = -5/1.4 m/s²
Now we can use Newton's second law:
F = m * a
Substituting the values:
F = (6 kg) * (-5/1.4 m/s²) ≈ -21.43 N
Since we are finding the magnitude, the average net force on the box is approximately 21.43 N.
b. By adding a 3 kg block inside the box, the total mass becomes 9 kg. To find the time it takes for the heavier box to stop, we can use the same formula as before.
Given:
Mass of the box (m) = 9 kg
Initial speed (u) = 5 m/s
Final speed (v) = 0 m/s (the box comes to a stop)
Using the equation of motion:
v = u + at
Rearranging the equation:
a = (v - u) / t
Substituting the given values:
0 = 5 m/s + a * t
Rearranging the equation:
a = -u / t
Substituting the values:
-5 m/s = -u / t
Now we can solve for t:
t = -u / (-5 m/s)
Substituting the given initial speed (u = 5 m/s):
t = -5 m/s / (-5 m/s) = 1 s
Therefore, it will take 1 second for the heavier box to stop.
To find the answers to these questions, we first need to understand the concept of net force and the relationship between net force, mass, and acceleration.
Net force represents the overall force acting on an object, taking into account the combined effect of all the individual forces acting on it. It can be calculated using Newton's second law of motion, which states that the net force applied on an object is equal to the product of its mass and acceleration (F = ma).
Let's proceed with solving the given problems step by step:
a. To find the magnitude of the average net force on the box, we need to use the equation F = ma. Given that the box has a mass of 6 kg and comes to a stop after 1.4 seconds, we can calculate the average acceleration using the equation a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Since the box starts with an initial speed of 5 m/s and comes to a stop (final speed is 0 m/s), Δv = 0 - 5 = -5 m/s.
Now, substituting the values into the equation:
F = m * a
F = 6 kg * (-5 m/s^2)
F = -30 N
The magnitude of the average net force on the box is 30 N.
b. When a 3 kg block is placed in the box, the total mass of the system becomes 9 kg (6 kg original box + 3 kg block). We need to determine how long it takes for this heavier box to stop when launched with an initial speed of 5 m/s.
Similar to the previous question, we'll calculate the average acceleration of the system using the equation a = Δv / Δt. However, this time Δv is 0 - 5 = -5 m/s, and we need to find Δt.
Rearranging the equation a = Δv / Δt, we get Δt = Δv / a.
Inserting the values we have:
Δt = -5 m/s / a
To find the acceleration (a) for the new system, we can use Newton's second law with the total mass (m = 9 kg) and solve for a:
F = m * a
-30 N = 9 kg * a
a = -30 N / 9 kg
a ≈ -3.33 m/s^2
Now, substituting the acceleration into the equation:
Δt = -5 m/s / (-3.33 m/s^2)
Δt ≈ 1.50 s
Therefore, it takes approximately 1.50 seconds for the heavier box (total mass 9 kg) to come to a stop when launched with an initial speed of 5 m/s.
a = change in v/ change in t = -5/1.4 = -3.57 m/s^2
so
F = m a = 6(-3.57) = - 21.4 Newtons
The friction force will be larger because proportional to the normal force (eight)
F = -21.4 (9/6) = -32.1 Newtons friction force
F = m a
a = -32.1 / 9 = 3.57 m/s^2
so same time, 1.4 seconds
(the mass and the retarding force go u in the same proportion)