A 150 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 420 N. For the first 21 m the floor is frictionless, and for the next 21 m the coefficient of friction is 0.30. What is the final speed of the crate?
Work put into the crate: force*distance=420*42 Joules
Work used by friction=150g*.3*21
KEfinal= workputintocrate-workusedonFriction.
velocityfinal= sqrt(2*KEfinal/mass)
11.51
To find the final speed of the crate, we need to calculate the acceleration of the crate in both portions of the floor and use the equations of motion to find its final speed.
First, let's consider the portion of the floor that is frictionless. In this case, there is no friction force opposing the motion of the crate. So, the only force acting on the crate is the applied horizontal force of 420 N.
To calculate the acceleration on the frictionless floor, we use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration:
F_net = m * a
In this case, the net force is the applied force F_applied, and the mass is the mass of the crate m. Therefore, we have:
F_applied = m * a (Equation 1)
Substituting the values given in the problem, we have:
420 N = 150 kg * a
Now, we can solve for the acceleration:
a = 420 N / 150 kg
a = 2.8 m/s^2
So, the acceleration of the crate on the frictionless floor is 2.8 m/s^2.
Next, let's consider the portion of the floor with a coefficient of friction of 0.30. In this case, there will be a friction force opposing the motion of the crate. The friction force can be calculated using the formula:
f_friction = coefficient of friction * normal force
The normal force is equal to the weight of the crate, which can be calculated as:
normal force = m * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values, we have:
normal force = 150 kg * 9.8 m/s^2
normal force = 1470 N
Now, we can calculate the friction force:
f_friction = 0.30 * 1470 N
f_friction = 441 N
The friction force acts in the opposite direction of the applied force, so the net force can be calculated as:
F_net = F_applied - f_friction
F_net = 420 N - 441 N
F_net = -21 N
Since the net force is negative, it means there is a deceleration in this portion of the floor. We can calculate the magnitude of the deceleration using Newton's second law again:
F_net = m * a
Substituting the values, we have:
-21 N = 150 kg * a
Solving for the acceleration:
a = -21 N / 150 kg
a = -0.14 m/s^2
So, the acceleration of the crate on the floor with a coefficient of friction of 0.30 is -0.14 m/s^2.
Now, we can use the equations of motion to find the final speed of the crate.
First, let's find the time taken to traverse each portion of the floor.
For the frictionless portion, we can use the following equation:
v = u + a * t (Equation 2)
where v is the final speed, u is the initial speed (which is zero since the crate starts from rest), a is the acceleration, and t is the time taken.
Substituting the known values, we have:
v = 0 + 2.8 m/s^2 * t
We can rearrange Equation 2 to solve for t:
t = v / (2.8 m/s^2)
Next, we'll use the equation of motion that relates displacement, initial velocity, final velocity, acceleration, and time:
s = ut + (1/2) * a * t^2 (Equation 3)
For the frictionless portion, the displacement s is given as 21 m, and the initial velocity u is zero. Substituting these values into Equation 3, we have:
21 m = 0 + (1/2) * 2.8 m/s^2 * t^2
Rearranging and solving for t^2:
t^2 = (2 * 21 m) / (2.8 m/s^2)
t^2 = 30 s^2
Now, we can solve for t:
t = sqrt(30 s^2)
t ≈ 5.48 s
So, it takes approximately 5.48 seconds to traverse the frictionless portion of the floor.
Now, let's calculate the time taken to traverse the portion with a coefficient of friction of 0.30.
Using the same equation (Equation 2), we have:
v = 0 + (-0.14 m/s^2) * t
Again, rearranging Equation 2 to solve for t:
t = v / (-0.14 m/s^2)
Next, using Equation 3 with a displacement of 21 m and initial velocity of 0, we have:
21 m = 0 + (1/2) * (-0.14 m/s^2) * t^2
Simplifying:
42 m = -0.07 m/s^2 * t^2
Solving for t^2:
t^2 = (42 m) / (-0.07 m/s^2)
t^2 ≈ -600 s^2
Since the time cannot be negative, it implies that the crate does not have enough applied force to traverse the portion with friction. This means that the crate will stop sliding on this portion of the floor.
Therefore, the final speed of the crate is 0 m/s.