In figure p4.36, m1 = 10kg and m2 = 4.0kg. the coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) if the system is released from rest, what will its acceleration be? (b) if the system is set in motion with m2 moving downward, what will be the acceleration of the system?
b.) M2g- μk M1g = (M1+ M2) a1
A1= (M2g- μk M1g )/(M1+ M2)
A1= ((4.0 kg) (9.8) – (0.30) (9.8) (10) )/(10+4)
A1= 0.7 m/s2
Well, let me first put on my clown shoes and get ready to entertain you with some physics humor!
(a) If the system is released from rest, it means it will finally get to go on that long-awaited vacation! But before it does, let's calculate the acceleration. We need to consider the forces acting on the system. The force of friction will oppose the motion, so we will subtract it from the force pulling the system:
Friction Force = (Coefficient of Static Friction) × (Normal Force)
The normal force is the force exerted by the surface, which is equal to the weight of m1:
Normal Force = (mass of m1) × (acceleration due to gravity)
Now, using Newton's second law (F = ma), we have:
(mass of m1) × (acceleration) - (Friction Force) = (mass of m1) × (acceleration)
Rearranging the equation, we get:
(mass of m1) × (acceleration) = (Friction Force)
Plugging in the values, we have:
(10 kg) × (acceleration) = (0.50) × [(10 kg) × (9.8 m/s^2)]
Simplifying the equation, we find:
(10 kg) × (acceleration) = 49 N
Dividing both sides by 10 kg, we get:
Acceleration = 4.9 m/s^2
So, the acceleration of the system when released from rest is 4.9 m/s^2. It's ready for a speedy getaway!
(b) Now, if m2 is moving downward and dragging the system along like a stubborn friend, let's find the acceleration. We'll have to consider both static and kinetic friction.
The force of static friction will be:
(Friction Force) = (Coefficient of Static Friction) × (Normal Force)
The force of kinetic friction will be:
(Friction Force) = (Coefficient of Kinetic Friction) × (Normal Force)
The normal force is the same as before, equal to the weight of m1:
Normal Force = (mass of m1) × (acceleration due to gravity)
Using Newton's second law once again, we have:
(Force pulling the system) - (Friction Force) = (mass of m1 + m2) × (acceleration)
Plugging in the values and rearranging the equation, we find:
[(mass of m1 + m2) × (acceleration)] = (Force pulling the system) - (Friction Force)
However, we have to consider that the friction force will reach a maximum value, equal to:
(Friction Force) = (Coefficient of Kinetic Friction) × (Normal Force)
So, [(mass of m1 + m2) × (acceleration)] = (Force pulling the system) - [(Coefficient of Kinetic Friction) × (Normal Force)]
With the given values, we can find the acceleration. I'll let you do the math and enjoy the joyride of solving equations!
Remember, laughter is the best way to accelerate through tough problems!
To find the acceleration of the system, we need to apply Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration.
(a) If the system is released from rest, we assume that there is no external force acting on the system. The only forces acting on the system are the force of gravity and the frictional force. The force of gravity acting on m1 is given by:
F_gravity1 = m1 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The frictional force between m1 and the surface can be found using the coefficient of static friction:
F_friction = μ_s * F_normal
where μ_s is the coefficient of static friction and F_normal is the normal force, which is equal to the force of gravity acting on m1 (F_normal = F_gravity1).
Substituting the values, the frictional force is:
F_friction = 0.50 * F_gravity1 = 0.50 * m1 * g
The net force acting on the system is equal to the force of gravity minus the frictional force:
F_net = F_gravity1 - F_friction
Using Newton's second law, we can write:
F_net = (m1 + m2) * a
Substituting the values, the equation becomes:
F_gravity1 - F_friction = (m1 + m2) * a
Simplifying the equation, we have:
m1 * g - 0.50 * m1 * g = (m1 + m2) * a
Rearranging the equation for acceleration:
a = (m1 * g - 0.50 * m1 * g) / (m1 + m2)
Substituting the given values, we can calculate the acceleration:
m1 = 10 kg
m2 = 4.0 kg
μ_s = 0.50
g = 9.8 m/s^2
a = (10 kg * 9.8 m/s^2 - 0.50 * 10 kg * 9.8 m/s^2) / (10 kg + 4.0 kg)
a = (98 N - 49 N) / 14 kg
a = 49 N / 14 kg
a ≈ 3.5 m/s^2
Therefore, the acceleration of the system when it is released from rest is approximately 3.5 m/s^2.
(b) If the system is set in motion with m2 moving downward, the frictional force between m1 and the surface changes to the kinetic friction force.
The frictional force can be calculated using the coefficient of kinetic friction:
F_friction = μ_k * F_normal
where μ_k is the coefficient of kinetic friction. Substituting the given values, we have:
F_friction = 0.30 * F_gravity1 = 0.30 * m1 * g
The net force acting on the system is again equal to the force of gravity minus the frictional force:
F_net = F_gravity1 - F_friction
Using Newton's second law, we can write:
F_net = (m1 + m2) * a
Substituting the values, the equation becomes:
F_gravity1 - F_friction = (m1 + m2) * a
Simplifying the equation, we have:
m1 * g - 0.30 * m1 * g = (m1 + m2) * a
Rearranging the equation for acceleration:
a = (m1 * g - 0.30 * m1 * g) / (m1 + m2)
Using the same values as before for m1, m2, and g:
a = (10 kg * 9.8 m/s^2 - 0.30 * 10 kg * 9.8 m/s^2) / (10 kg + 4.0 kg)
a = (98 N - 29.4 N) / 14 kg
a = 68.6 N / 14 kg
a ≈ 4.9 m/s^2
Therefore, the acceleration of the system when m2 is moving downward is approximately 4.9 m/s^2.
To find the acceleration of the system, we need to consider two scenarios: when the system is released from rest and when the system is set in motion with m2 moving downward. Let's compute the acceleration for each case:
(a) When the system is released from rest:
In this scenario, we need to determine whether the static friction between m1 and the surface is enough to prevent m1 and m2 from moving. We compare the static friction force (fs) to the maximum static friction force (fs_max) using the formula:
fs_max = μs * N,
where μs is the coefficient of static friction and N is the normal force.
The normal force N acting on m1 is equal to its weight, which is mg1, where g is the acceleration due to gravity.
The maximum static friction force fs_max acts in the opposite direction to the applied force (m2g). Considering that the system is in equilibrium before being released, fs_max is equal in magnitude but opposite in direction to the applied force, so fs_max = m2g.
Comparing the actual static friction force fs to fs_max:
If the static friction force (fs) is equal to fs_max, m1 and m2 will not move, and the acceleration of the system will be 0. However, if fs_max > fs, then we need to consider the kinetic friction force fk.
fs = μs * N,
where N is the magnitude of the normal force, and N = mg1 in this case. Therefore:
fs = μs * mg1.
If fs < fs_max, then the system will move, and the resultant force will be in the direction of m2g (downward). The net force acting on the system will be:
net force = m2g - fs.
Using Newton's second law (F = ma), we can set up the following equation:
m1a = m2g - fs.
Rearranging and substituting fs = μs * mg1:
m1a = m2g - μs * mg1.
Simplifying:
a = (m2g - μs * mg1) / m1.
Substituting the given values:
m1 = 10 kg, m2 = 4.0 kg, μs = 0.50, g = 9.8 m/s^2:
a = (4.0 * 9.8 - 0.50 * 10 * 9.8) / 10
= (39.2 - 0.50 * 98) / 10
= (39.2 - 49) / 10
= -9.8 / 10
= -0.98 m/s^2.
Therefore, the acceleration of the system when released from rest is -0.98 m/s^2 in the direction opposite to m2.
(b) When the system is set in motion with m2 moving downward:
In this scenario, the kinetic friction force (fk) between m1 and the surface is relevant. The kinetic friction force fk is given by:
fk = μk * N,
where μk is the coefficient of kinetic friction.
Since the direction of motion of m1 and m2 is now downward, the force acting on the system is m2g. Therefore, the net force acting on the system is:
net force = m2g - fk.
Using Newton's second law (F = ma), and substituting fk = μk * N, we can set up the following equation:
(m1 + m2) * a = m2g - μk * mg1.
Simplifying and substituting the given values:
m1 = 10 kg, m2 = 4.0 kg, μk = 0.30, g = 9.8 m/s^2:
(10 + 4.0) * a = 4.0 * 9.8 - 0.30 * 10 * 9.8
14.0a = 39.2 - 2.94
14.0a = 36.26
a = 36.26 / 14.0
a ≈ 2.59 m/s^2.
Therefore, the acceleration of the system when set in motion with m2 moving downward is approximately 2.59 m/s^2 in the direction of m2.