A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 420 N. For the first 13 m the floor is frictionless, and for the next 13 m the coefficient of friction is 0.35. What is the final speed of the crate?
The final speed of the crate is 8.7 m/s.
Well, well, well, looks like we have a crate in motion! Let's see if we can unravel this puzzle.
First up, we're dealing with two different sections of the floor. The first 13 m is as slippery as a buttered-up banana peel, so there's no friction to slow down our crate.
Now, for the next 13 m, we've got a coefficient of friction of 0.35. That's like having a persistent little gremlin trying to put the brakes on our crate's speedy adventure.
To find the final speed of the crate, we'll need to break this problem down into two parts. Are you ready for some math magic?
Part 1: The frictionless section
Since there's no friction, the net force acting on the crate is simply the pulling force of 420 N. So, we can use good old Newton's second law (F = ma) to find the crate's acceleration (a) in this section.
420 N = (110 kg) * a
Solving for a, we get:
a = 420 N / 110 kg
a = 3.82 m/s^2
Now, we can use a little kinematics to find the velocity (v) at the end of the first section. The equation v^2 = u^2 + 2as comes to the rescue!
u = initial velocity = 0 m/s
a = acceleration = 3.82 m/s^2
s = distance traveled = 13 m
Plugging in the numbers, we get:
v^2 = (0 m/s)^2 + 2 * (3.82 m/s^2) * (13 m)
v^2 = 98.68 m^2/s^2
Taking the square root, we find:
v ≈ 9.93 m/s (rounded to 2 decimal places)
Part 2: The friction-full section
In this section, we need to deal with the gremlin-like friction force. The friction force can be found using the equation F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force.
In this case, the normal force is equal to the weight of the crate, since the floor is horizontal and there are no other vertical forces acting on it.
F_normal = m * g
F_normal = (110 kg) * (9.8 m/s^2)
F_normal = 1078 N
Now, we can find the friction force:
F_friction = μ * F_normal
F_friction = 0.35 * 1078 N
F_friction ≈ 377 N
With the friction force determined, we can find the net force acting on the crate in this section:
Net force = Pulling force - Friction force
Net force = 420 N - 377 N
Net force = 43 N
Applying Newton's second law (F = ma) once again:
43 N = (110 kg) * a
Solving for a, we get:
a = 43 N / 110 kg
a ≈ 0.39 m/s^2
Now, we can calculate the distance (s) covered in this section:
a = (v^2 - u^2) / (2s)
0.39 m/s^2 = (v^2 - (9.93 m/s)^2) / (2 * 13 m)
Solving for v^2, we find:
v^2 ≈ (0.39 m/s^2) * (2 * 13 m) + (9.93 m/s)^2
v^2 ≈ (0.39 m/s^2) * 26 m + (9.93 m/s)^2
v^2 ≈ 21.54 m^2/s^2 + 98.68 m^2/s^2
v^2 ≈ 120.22 m^2/s^2
Taking the square root, we find:
v ≈ 10.97 m/s (rounded to 2 decimal places)
So, my friend, drumroll please...the final speed of that crate is approximately 10.97 m/s. It went from a smooth ride to feeling like it had a mischievous gremlin riding along. But hey, at least it made it to the end!
To find the final speed of the crate, we need to calculate the total work done on the crate and then use the work-energy principle.
First, let's calculate the work done during the first 13 m when the floor is frictionless. Since there is no friction, the only force acting on the crate is the applied force of 420 N. The work done by a constant force is given by the equation:
Work = Force × Distance × Cos(θ)
In this case, since the force and distance are in the same direction, the angle θ is 0. Therefore, the work done during the first 13 m is:
Work1 = 420 N × 13 m × Cos(0)
= 5460 J
Next, let's calculate the work done during the next 13 m when the coefficient of friction is 0.35. The force of friction can be calculated using the equation:
Frictional force = coefficient of friction × Normal force
The normal force is equal to the weight of the crate, which can be calculated using the equation:
Weight = mass × acceleration due to gravity
Weight = 110 kg × 9.8 m/s^2
= 1078 N
Therefore, the frictional force is:
Frictional force = 0.35 × 1078 N
= 377.3 N
The work done by the frictional force is given by:
Work2 = Frictional force × Distance × Cos(180)
Since the force and displacement are in opposite directions, the angle θ is 180 degrees. Therefore, the work done during the next 13 m is:
Work2 = 377.3 N × 13 m × Cos(180)
= -4878.6 J (negative because the displacement and force are opposite in direction)
Now, let's find the total work done on the crate:
Total work = Work1 + Work2
= 5460 J + (-4878.6 J)
= 582.6 J
According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. Therefore, we can equate the total work to the change in kinetic energy:
Total work = Change in kinetic energy
Since the crate starts from rest, its initial kinetic energy is 0:
Change in kinetic energy = Total work - Initial kinetic energy
= Total work - 0
= Total work
Therefore, the change in kinetic energy is 582.6 J.
Finally, we can calculate the final speed of the crate using the equation:
Final kinetic energy = (1/2) × mass × final velocity^2
Since the crate starts from rest, its initial kinetic energy is 0. Therefore:
Final kinetic energy = Change in kinetic energy
= 582.6 J
Substituting the values into the equation and solving for the final velocity:
(1/2) × 110 kg × final velocity^2 = 582.6 J
final velocity^2 = (582.6 J) / [(1/2) × 110 kg]
= 10.57 m^2/s^2
final velocity = sqrt(10.57 m^2/s^2)
= 3.25 m/s
Therefore, the final speed of the crate is 3.25 m/s.
To find the final speed of the crate, we need to break down the problem into two parts: the frictionless region and the region with friction.
1. Frictionless region (13 m):
Since there is no friction, the only force acting on the crate is the pulling force of 420 N. We can use Newton's second law to find the acceleration of the crate:
F = ma
420 N = 110 kg * a
a = 3.82 m/s^2
Using the kinematic equation, we can find the final velocity in this region:
v^2 = u^2 + 2as
u = 0 m/s (initial velocity, as the crate starts from rest)
v^2 = 0^2 + 2 * 3.82 m/s^2 * 13 m
v^2 = 98.92 m^2/s^2
v ≈ 9.94 m/s
2. Region with friction (13 m):
In this region, we need to consider the frictional force opposing the motion. The frictional force is given by:
frictional force = coefficient of friction * normal force
The normal force can be calculated as the weight of the crate:
weight = mass * gravity
weight = 110 kg * 9.8 m/s^2
weight ≈ 1078 N
Now, let's calculate the frictional force:
frictional force = 0.35 * 1078 N
frictional force ≈ 377.3 N
The net force acting on the crate in this region can be calculated as:
net force = applied force - frictional force
net force = 420 N - 377.3 N
net force ≈ 42.7 N
Again, using Newton's second law, we can find the acceleration in this region:
42.7 N = 110 kg * a
a ≈ 0.39 m/s^2
Using the same kinematic equation as before, we can find the final velocity in this region:
v^2 = u^2 + 2as
u ≈ 9.94 m/s (initial velocity in this region is the final velocity from the previous region)
v^2 = 9.94^2 + 2 * 0.39 m/s^2 * 13 m
v^2 = 124.03 m^2/s^2
v ≈ 11.13 m/s
Therefore, the final speed of the crate is approximately 11.13 m/s.