A 40 N force is applied at an angle of 30o to the horizontal to pull a 6 kg box along the floor. If the coefficient of friction between the box and floor is 0.3.The box initially at rest moves a horizontal distance of 5m. What is the speed of the box when it reaches 5m?
No idea how to do it
To determine the speed of the box when it reaches 5m, we first need to calculate the acceleration of the box using Newton's second law of motion. Here are the steps to solve this problem:
Step 1: Calculate the gravitational force acting on the box.
The force of gravity (weight) acting on the box can be calculated using the formula:
Weight = mass x gravitational acceleration (W = m x g)
Given:
Mass of box (m) = 6 kg
Gravitational acceleration (g) = 9.8 m/s^2
Weight = 6 kg x 9.8 m/s^2 = 58.8 N
Step 2: Resolve the applied force into horizontal and vertical components.
The applied force is given as 40 N at an angle of 30 degrees to the horizontal. We need to determine the horizontal component of this force, which will affect the acceleration of the box.
Horizontal force (F_h) = Applied force (F) x Cosine(angle)
F_h = 40 N x Cos(30°)
F_h = 40 N x √3/2
F_h = 20√3 N
Step 3: Calculate the force of friction.
The force of friction opposing the motion can be calculated using the formula:
Force of friction (F_friction) = Coefficient of friction (μ) x Normal force
Given:
Coefficient of friction (μ) = 0.3
Normal force (N) = Weight of the box (W)
F_friction = μ x N
F_friction = 0.3 x 58.8 N
F_friction = 17.64 N
Step 4: Calculate the net force acting on the box.
The net force can be determined by subtracting the force of friction from the horizontal component of the applied force.
Net force (F_net) = F_h - F_friction
F_net = 20√3 N - 17.64 N
F_net = 2.36 N
Step 5: Calculate the acceleration of the box.
The acceleration (a) of the box can be calculated using Newton's second law of motion:
F_net = m x a
2.36 N = 6 kg x a
Solving for a:
a = 2.36 N / 6 kg
a ≈ 0.393 m/s^2
Step 6: Use kinematic equations to determine the speed of the box.
We know that the box starts from rest, so the initial velocity (u) is 0 m/s. The final velocity (v) is what we need to find, and the distance traveled (s) is given as 5m.
Using the equation:
v^2 = u^2 + 2as
Substituting the known values:
v^2 = 0 + 2 x 0.393 m/s^2 x 5 m
v^2 = 3.93 m^2/s^2
Taking the square root of both sides:
v = √3.93 m/s
v ≈ 1.98 m/s
Therefore, the speed of the box when it reaches 5m is approximately 1.98 m/s.