1. A 50.0 Lb box is to be moved along a rough floor at a constant velocity. The coefficient of the kinetic friction is 0.300.
A. What force must you exert if you push downward on the box?
B. What force must you exert if you pull upward on the box?
C. Which is the better way to move the box?
2. Find :
A. The magnitude of the acceleration of the system shown if the coefficient of kinetic friction B is 0.300, Coefficient if kinetic friction A is 0.200, MB=3.00 kg, MA= 5.00kg.
B. The velocity of block A at 0.500s?
3. A 10.0 kg package slides down on inclined mail chute 15.0m long. The top of the chute is 6.00 above the floor. What is the speed of package at the button of the chute if the coefficient of kinetic friction is 0.300?
1. Neither. Pushing down does nothing, pulling up moves it vertically.
2. Amazing Kreskin abilities fading...
3. The angle is 23.6o so Normal force is 10*9.8*cos23.6 and the friction force is .3 * Normal force. Friction force will give you work lost to friction (Fd where d =15). Finally take initial PE = mgh minus work lost to friction and set it equal to final KE (.5 m v^2).
1. A. To find the force you must exert if you push downward on the box, you can use the equation for the force of friction:
F_friction = μ_k * N
where μ_k is the coefficient of kinetic friction and N is the normal force. The normal force can be calculated as the weight of the box:
N = m * g
where m is the mass of the box and g is the acceleration due to gravity. Substituting this into the equation for friction, we get:
F_friction = μ_k * m * g
In this case, the mass of the box is given as 50.0 lb, which can be converted to kilograms by multiplying by the conversion factor 0.4536 kg/lb. The acceleration due to gravity is approximately 9.8 m/s^2. Plug in the values:
F_friction = 0.300 * 50.0 lb * 0.4536 kg/lb * 9.8 m/s^2
Calculating this will give you the force you need to exert if you push downward on the box.
B. To find the force you must exert if you pull upward on the box, you can use the same equation for the force of friction:
F_friction = μ_k * N
However, in this case, the normal force will be different. When you pull upward on the box, the normal force will be reduced by the force you exert. So the normal force will be:
N = m * g - F_pull
where F_pull is the force you exert when you pull upward on the box. Again substituting into the equation for friction, we get:
F_friction = μ_k * (m * g - F_pull)
Solving this equation will give you the force you need to exert if you pull upward on the box.
C. The better way to move the box depends on the forces involved. If the force you need to exert when pushing downward is smaller than the force you need to exert when pulling upward, then pushing downward would be the better way. Conversely, if the force when pulling upward is smaller, then pulling upward would be the better way. You can compare the magnitudes of the forces found in parts A and B to determine which is smaller.
2. A. To find the magnitude of the acceleration of the system, we need to consider the forces acting on it. The system consists of two blocks connected by a rope, and friction acts between the blocks and the surface they are on. The force diagram for the system can be drawn as follows:
-----> (Tension force) -------> (Block A)
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| |
| <------ (Friction A) <------ |
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| |
| -----> (Friction B) ------> | <--- (Block B)
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| |
Applying Newton's second law to each block gives us:
For block A: MA * a = Tension force - Friction A
For block B: MB * a = Friction B
The force of friction can be calculated using the equation:
Friction = μ_k * N
where μ_k is the coefficient of kinetic friction and N is the normal force. For block A, the normal force is MA * g, and for block B, the normal force is MB * g, where g is the acceleration due to gravity. Substituting these values into the friction equations, we get:
Friction A = μ_k * (MA * g)
Friction B = μ_k * (MB * g)
Substituting these equations into the block equations, we get:
MA * a = Tension force - μ_k * (MA * g)
MB * a = μ_k * (MB * g)
Rearranging these equations, we can solve for the acceleration:
a = (Tension force - μ_k * (MA * g)) / MA
a = μ_k * (MB * g) / MB
Now, we can substitute the given values:
MA = 5.00 kg
MB = 3.00 kg
μ_k = 0.300
g = 9.8 m/s^2
and solve for the magnitude of the acceleration, a.
B. To find the velocity of block A at 0.500s, we need to use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the system starts from rest, so the initial velocity u is 0 m/s. We already have the acceleration, which we found in part A. The time t is given as 0.500s.
Substituting these values into the equation, we can solve for the velocity, v.
1. A. To determine the force required to push the box downward on the rough floor, we first need to calculate the frictional force acting on the box. The formula to calculate the frictional force is given by:
Frictional force = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the box, which can be calculated using the formula:
Weight = mass * acceleration due to gravity
Since the problem states that the box weighs 50.0 lb, which is a unit of force, we need to convert it to mass in order to calculate the weight. The conversion factor is 1 lb = 0.4536 kg.
So, the mass of the box (m) can be calculated as:
m = weight / acceleration due to gravity = (50.0 lb) / (32.2 ft/s² / 3.281 ft/m) * (0.4536 kg / 1 lb) ≈ 22.7 kg
The frictional force (Ff) can be calculated as:
Ff = coefficient of kinetic friction * normal force = 0.300 * (mass * acceleration due to gravity) = 0.300 * (22.7 kg * 9.8 m/s²) ≈ 66.6 N
Since the box is moving at a constant velocity, the pushing force must be equal in magnitude and opposite in direction to the frictional force. Therefore, the force you must exert to push downward on the box is approximately 66.6 N.
B. To determine the force required to pull upward on the box, we can use the same calculation as in part A, since the frictional force will be the same regardless of the direction.
Therefore, the force you must exert to pull upward on the box is also approximately 66.6 N.
C. Since the force required to push downward on the box is the same as the force required to pull upward, neither method is better in terms of the magnitude of force required. However, depending on the situation, one method may be more convenient than the other (e.g., if you are standing above the box, pushing downward may be easier). The choice of method would also depend on other factors like the weight distribution and stability of the box.
2. A. To find the magnitude of the acceleration of the system, we need to consider the forces acting on the blocks. The key forces at play here are the force of gravity (mg) and the force of friction (Ff) due to the coefficients of kinetic friction.
The formula to calculate the net force on each block is:
Net force (block A) = mA * acceleration (block A)
Net force (block B) = mB * acceleration (block B)
The force of gravity acting on each block can be calculated as:
Force of gravity (block A) = mA * acceleration due to gravity
Force of gravity (block B) = mB * acceleration due to gravity
The force of friction acting on each block can be calculated as:
Force of friction (block A) = coefficient of kinetic friction A * normal force (block A)
Force of friction (block B) = coefficient of kinetic friction B * normal force (block B)
The normal force on each block can be calculated as:
Normal force (block A) = mA * acceleration due to gravity
Normal force (block B) = mB * acceleration due to gravity
By equating the net forces and using the given values, we can solve for the acceleration:
(mA * acceleration) - (coefficient of kinetic friction A * mA * acceleration due to gravity) = mA * acceleration
...
(mB * acceleration) - (coefficient of kinetic friction B * mB * acceleration due to gravity) = -mA * acceleration
Simplifying the equations further, we get:
acceleration = (coefficient of kinetic friction B * mB * acceleration due to gravity) / (mA + mB) = (0.300 * 3.00 kg * 9.8 m/s²) / (5.00 kg + 3.00 kg) ≈ 1.76 m/s²
Therefore, the magnitude of the acceleration of the system is approximately 1.76 m/s².
B. To find the velocity of block A at 0.500 s, we need to use the formula for constant acceleration:
velocity (block A) = initial velocity (block A) + acceleration (block A) * time
Since block A starts from rest, the initial velocity is 0 m/s.
velocity (block A) = 0 m/s + (acceleration (block A) * time) = 1.76 m/s² * 0.500 s ≈ 0.88 m/s
Therefore, the velocity of block A at 0.500 s is approximately 0.88 m/s.
3. To find the speed of the package at the bottom of the inclined mail chute, we need to consider the forces acting on the package. The key forces at play here are the force of gravity (mg), the normal force (N), and the force of friction (f) due to the coefficient of kinetic friction.
The force of gravity acting on the package can be calculated as:
Force of gravity (package) = mass (package) * acceleration due to gravity = 10.0 kg * 9.8 m/s² ≈ 98.0 N
The normal force acting on the package can be calculated as:
Normal force (package) = weight (package) * cos(angle of incline) = 98.0 N * cos(arctan(6.00 m / 15.0 m)) ≈ 83.1 N
The force of friction acting on the package can be calculated as:
Force of friction (package) = coefficient of kinetic friction * normal force (package) = 0.300 * 83.1 N ≈ 24.9 N
The net force acting on the package can be calculated as:
Net force (package) = force of gravity (package) - force of friction (package) = 98.0 N - 24.9 N ≈ 73.1 N
Since the force of friction acts opposite to the direction of motion, the net force is responsible for the acceleration of the package. However, at the bottom of the inclined chute, the package will have reached a constant speed. Therefore, the magnitude of the net force is equal to zero.
Setting the net force equal to zero, we can solve for the speed at the bottom of the chute:
0 = force of gravity (package) - force of friction (package) - mass (package) * acceleration = 98.0 N - 24.9 N - (10.0 kg * acceleration)
Simplifying the equation, we find:
acceleration = (98.0 N - 24.9 N) / 10.0 kg ≈ 7.51 m/s²
Now, we can find the speed using the equation:
speed = √(2 * acceleration * distance)
distance = 15.0 m (given)
speed = √(2 * 7.51 m/s² * 15.0 m) ≈ 11.0 m/s
Therefore, the speed of the package at the bottom of the chute is approximately 11.0 m/s.