A box of books weighing 314 N is shoved across the floor by a force of 475 N exerted downward at an angle of 35° below the horizontal.
(a) If μk between the box and the floor is 0.53, how long does it take to move the box 5.00 m starting from rest?
What is the maximum coefficient of kinetic friction between the box and the floor that allows the box to move from this applied force.
Wb = 314 N.
m*g = 314
m = 314/g = 314/9.8=32 kg.=mass of box.
Fb = 314N[0o].
Fp = 314*sin(0) = 0 = Force parallel to floor.
Fv = 314*cos(0)+475*sin35 = 586 N. =
Force perpendicular to floor.
Fk = u*Fv = 0.53 * 586 = 310.6 N. = Force of kinetic friction.
a. Fap*cos35-Fp-Fk = m*a
475*cos35-0-310.6 = 32*a
389-310.6 = 32a
78.4 = 32a
a = 2.45m/s^2.
d = 0.5a*t^2 = 5 m.
0.5*2.45t^2 = 5
1.225t^2 = 5
t^2 = 4.082
t = 2.02 s.
Well, it looks like the box is quite a handful! Let's see if we can solve this with a pinch of humor:
(a) To calculate the time it takes for the box to move, we need to consider the net force acting on it. In this case, the force exerted downward at an angle of 35° is the driving force, while the frictional force is in the opposite direction. So, the net force is given by:
Net Force = Applied Force - Frictional Force
The applied force is 475 N, and the frictional force can be calculated using the equation:
Frictional Force = coefficient of kinetic friction * Normal Force
The normal force is the force exerted by the floor to support the box's weight, which is equal to its weight. So, the frictional force is:
Frictional Force = 0.53 * 314 N
Now, using the net force equation, we can find the acceleration:
Net Force = Mass * Acceleration
Solving for acceleration, we get:
Acceleration = Net Force / Mass
The mass of the box is given by its weight divided by the acceleration due to gravity:
Mass = Weight / Gravity
Now, we can use the equations of motion to find the time it takes for the box to move a distance of 5.00 m, starting from rest:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
Since the initial velocity is zero, the equation simplifies to:
Distance = (1/2) * Acceleration * Time^2
Rearranging the equation, we can find the time:
Time = sqrt((2 * Distance) / Acceleration)
Now, let's plug in the numbers and calculate the time!
As for the maximum coefficient of kinetic friction that allows the box to move, it depends on the net force. The box will start moving when the applied force is greater than the frictional force:
Applied Force > Frictional Force
Let's calculate the maximum coefficient of kinetic friction using this equation!
Alright, enough with the calculations! I hope I didn't give your brain too much of a workout. But trust me, once you solve this problem, you'll feel like you could write a whole physics comedy routine!
To find the time it takes to move the box, we can use the equation of motion:
d = vit + (1/2)at²
where:
d = distance (5.00 m)
vi = initial velocity (0 m/s) since the box starts from rest
t = time taken
We need to find the acceleration, a, which is given by:
a = Fnet / m
where:
Fnet = net force acting on the box
m = mass of the box (which can be calculated from weight)
The net force is the horizontal component of the applied force minus the force of friction, so:
Fnet = Fapplied - Ffriction
The horizontal component of the applied force is given by:
Fapplied = F * cos(θ)
where:
F = magnitude of the applied force (475 N)
θ = angle of the applied force (35°)
The force of friction can be found using:
Ffriction = μk * Fn
where:
μk = coefficient of kinetic friction (0.53)
Fn = normal force
In this case, the normal force is equal to the weight of the box:
Fn = weight of the box = mg
Given that the weight of the box is 314 N, we can equate it to mg:
314 = m * g
Now, we have all the necessary components to solve the problem. Let's calculate each step systematically:
Step 1: Calculate the mass of the box (m):
m = weight / g = 314 N / 9.81 m/s²
m ≈ 32 kg (approx.)
Step 2: Calculate the horizontal component of the applied force (Fapplied):
Fapplied = F * cos(θ)
Fapplied = 475 N * cos(35°)
Fapplied ≈ 389.36 N (approx.)
Step 3: Calculate the force of friction (Ffriction):
Ffriction = μk * Fn
Ffriction = 0.53 * 314 N
Ffriction ≈ 166.42 N (approx.)
Step 4: Calculate the net force (Fnet):
Fnet = Fapplied - Ffriction
Fnet ≈ 389.36 N - 166.42 N
Fnet ≈ 222.94 N (approx.)
Step 5: Calculate the acceleration (a):
a = Fnet / m
a ≈ 222.94 N / 32 kg
a ≈ 6.97 m/s² (approx.)
Step 6: Calculate the time taken (t):
Using the equation of motion:
d = vit + (1/2)at²
t² = (2d - 2vi*t) / a
t² = (2 * 5.00 m) / (6.97 m/s²)
t² ≈ 1.43 s² (approx.)
t ≈ √(1.43 s²)
t ≈ 1.20 s (approx.)
Therefore, it takes approximately 1.20 seconds to move the box a distance of 5.00 meters starting from rest.
Now let's find the maximum coefficient of kinetic friction that allows the box to move:
Given:
Fapplied = 475 N
θ = 35°
Using the same steps as above, we can calculate the force of friction (Ffriction) and solve for μk:
Step 1: Calculate the horizontal component of the applied force (Fapplied):
Fapplied = F * cos(θ)
Fapplied = 475 N * cos(35°)
Fapplied ≈ 389.36 N (approx.)
Step 2: Calculate the force of friction (Ffriction):
Ffriction = μk * Fn
Fn = weight of the box = mg = 314 N
Ffriction = μk * 314 N
We need to find the maximum coefficient of kinetic friction (μk) that allows the box to move:
Ffriction = μk * 314 N
Since the box is just about to move, the force of friction is at its maximum value:
FfrictionMax = μkMax * 314 N
Therefore, the maximum value for the coefficient of kinetic friction (μkMax) is given by:
μkMax = FfrictionMax / 314 N
μkMax = Fapplied / 314 N
Using the previously calculated value for Fapplied:
μkMax = 389.36 N / 314 N
μkMax ≈ 1.24 (approx.)
Therefore, the maximum coefficient of kinetic friction (μkMax) that allows the box to move from the applied force is approximately 1.24
To solve this problem, we need to break it down into two parts:
Part 1: Calculate the time it takes to move the box 5.00 m starting from rest.
Part 2: Determine the maximum coefficient of kinetic friction that allows the box to move from the applied force.
Part 1: Calculate the time it takes to move the box 5.00 m starting from rest.
To calculate the time it takes to move the box, we can use the equation of motion:
Δd = V₀t + (1/2)at²
Where:
Δd = displacement (5.00 m),
V₀ = initial velocity (0 m/s since the box starts from rest),
t = time taken to move the box,
a = acceleration.
To find the acceleration, we need to consider the net force acting on the box:
Net force = Applied force - Force of friction
The applied force is given as 475 N at an angle of 35° below the horizontal.
Force of friction = μk * Normal force
The normal force is equal to the weight of the box, which is 314 N.
Substituting these values into the equations:
Net force = 475 N * cos(35°) - (0.53 * 314 N)
Next, we can use Newton's second law of motion:
Net force = mass * acceleration
Rearranging the equation, we get:
Acceleration = Net force / mass = (475 N * cos(35°) - (0.53 * 314 N)) / mass
The mass is not given, but we can calculate it using the weight of the box:
Weight = mass * gravity
Rearranging the equation, we get:
Mass = Weight / gravity = 314 N / 9.8 m/s²
Now we can substitute the values of the net force and mass into the acceleration equation:
Acceleration = (475 N * cos(35°) - (0.53 * 314 N)) / (314 N / 9.8 m/s²)
Now that we have the acceleration, we can substitute it into the equation of motion to solve for time:
5.00 m = 0 m/s * t + (1/2) * [(475 N * cos(35°) - (0.53 * 314 N)) / (314 N / 9.8 m/s²)] * t²
Simplifying and solving for t will give us the time it takes to move the box 5.00 m starting from rest.
Part 2: Determine the maximum coefficient of kinetic friction that allows the box to move from the applied force.
To determine the maximum coefficient of kinetic friction, we need to consider the condition where the box just starts to move. In this case, the force of friction is equal to the maximum force of static friction, given by:
Maximum force of static friction = μs * Normal force
Since the box just starts to move, the static friction is the limiting factor. Therefore, the maximum coefficient of kinetic friction (μk) is equal to the coefficient of static friction (μs):
μk = μs
We do not have the value of μs in the problem, so we cannot determine the maximum coefficient of kinetic friction without additional information.