A box of 6kg is pulled to the right on horizontal surface by force of 20 N. The box experience a constant friction force of 3N during it's motion
20 - 3 = m a
17 = 6 a
a = 17 / 6
So what is the question? This seems to be the day for half worded questions.
What was its initial speed? How long to accelerate to whatever speed? Distance during acceleration?
Calculate normal force.
The normal force on the ground is m g = 6 * 9.81 Newtons
To determine the acceleration and net force acting on the box, we can use Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a).
Given:
Mass of the box (m) = 6 kg
Applied force (F_applied) = 20 N
Friction force (F_friction) = 3 N
First, let's find the net force:
F_net = F_applied - F_friction
F_net = 20 N - 3 N
F_net = 17 N
Now, we will use Newton's second law to find the acceleration:
F_net = m * a
17 N = 6 kg * a
Dividing both sides of the equation by 6 kg:
a = 17 N / 6 kg
a ≈ 2.83 m/s^2
So, the acceleration of the box is approximately 2.83 m/s^2 and the net force acting on the box is 17 N.