The figure shows four penguins that are being playfully pulled along very slippery (frictionless) ice by a curator. The masses of three penguins and the tension in two of the cords are m1 = 10.5 kg, m3 = 14.7 kg, m4 = 22.05 kg, T2 = 120 N, and T4 = 236 N. What is the mass of the second penguin (the mass m2 not given)?
m1 = 10.5 kg, m3 = 14.7 kg, m4 = 22.05 kg, T2 = 120 N, and T4 = 236 N.
m2=?
(m1+m2+m3+m4)a=T4
(m3+m4)a=T4-T2
(10.5+m2+14.7+22.05)a =236
(14.7+22.05)a=236-120
a=116/36.75 =3.16 m/s²
(10.5+m2+14.7+22.05)a =236
m2={236-(10.5+14.7+22.05)3.16}/3.16 = 27.43 kg
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To find the mass of the second penguin (m2), we can start by analyzing the forces acting on each penguin using Newton's laws of motion. Considering the figure, let's denote the masses of the penguins as follows:
m1 = 10.5 kg (given mass)
m2 = unknown
m3 = 14.7 kg (given mass)
m4 = 22.05 kg (given mass)
Next, let's analyze the tensions in the cords:
T2 = 120 N (given tension)
T4 = 236 N (given tension)
There are a few key concepts we need to consider:
1. The net force on each penguin is equal to the product of its mass and acceleration (F = ma).
2. The acceleration of each penguin is the same since they are all being pulled together.
3. The tension in each cord is the same since they are all connected, and the cords are assumed to be massless and frictionless.
Now, consider the forces acting on each penguin:
For m1:
- There is tension T2 pulling it forward.
- There is no other force acting on it.
- So, the net force on m1 is T2.
For m2:
- There is no tension or any other force acting on it.
- The net force on m2 is zero since it is not being pulled in any direction.
For m3:
- There is no tension or any other force acting on it.
- The net force on m3 is zero.
For m4:
- There is tension T4 pulling it backward.
- There is no other force acting on it.
- So, the net force on m4 is T4.
Since the net force on each penguin is equal to the product of its mass and acceleration, we can set up the following equations:
For m1: F1 = T2 = m1 * a
For m2: F2 = 0 = m2 * a
For m3: F3 = 0 = m3 * a
For m4: F4 = T4 = m4 * a
Since the acceleration (a) is the same for all penguins, we can write:
T2 = m1 * a
0 = m2 * a
0 = m3 * a
T4 = m4 * a
Now, let's substitute the given values into the equations:
T2 = m1 * a
120 N = 10.5 kg * a
0 = m2 * a
0 = m3 * a
T4 = m4 * a
236 N = 22.05 kg * a
Now, let's solve the equation T2 = m1 * a for acceleration (a):
a = T2 / m1
a = 120 N / 10.5 kg
a ≈ 11.43 m/s²
Since the acceleration (a) is the same for all penguins, we can use it to find the mass of the second penguin (m2):
0 = m2 * a
0 = m2 * 11.43 m/s²
Since the left side of the equation is zero, the mass of the second penguin (m2) must also be zero.
Therefore, the mass of the second penguin is zero.
To find the mass of the second penguin (m2), we need to use the concept of Newton's second law of motion and the fact that the ice is frictionless.
First, let's analyze the forces acting on each penguin:
1. For penguin 1 (m1): The only force acting on it is the tension in cord T2, pulling it to the right.
2. For penguin 2 (m2): We want to find its mass, so we don't know the forces acting on it directly.
3. For penguin 3 (m3): The forces acting on it are the tension in T2 pulling to the left and the tension in T4 pulling to the right.
4. For penguin 4 (m4): The only force acting on it is the tension in cord T4, pulling it to the left.
Now, let's write down the equations of motion for the system:
For penguin 1 (m1): T2 = m1 * a1
For penguin 2 (m2): Unknown
For penguin 3 (m3): T2 - T4 = m3 * a3
For penguin 4 (m4): T4 = m4 * a4
Since the tensions in cords T2 and T4 are known (T2 = 120 N and T4 = 236 N) and the masses m1, m3, and m4 are given (m1 = 10.5 kg, m3 = 14.7 kg, and m4 = 22.05 kg), we can solve for the acceleration of each penguin.
First, let's find the acceleration of penguin 1 (a1) using the equation T2 = m1 * a1:
a1 = T2 / m1
a1 = 120 N / 10.5 kg
a1 ≈ 11.43 m/s²
Next, let's find the acceleration of penguin 3 (a3) using the equation T2 - T4 = m3 * a3:
a3 = (T2 - T4) / m3
a3 = (120 N - 236 N) / 14.7 kg
a3 ≈ -9.69 m/s²
Finally, let's find the acceleration of penguin 4 (a4) using the equation T4 = m4 * a4:
a4 = T4 / m4
a4 = 236 N / 22.05 kg
a4 ≈ 10.69 m/s²
Since the system is in equilibrium (the penguins are not accelerating), the acceleration of penguin 2 (a2) is zero.
Now, let's find the mass of penguin 2 (m2) using the equation T2 = m2 * a2:
0 = m2 * 0
0 = m2
The mass of penguin 2 (m2) is 0 kg or it is not given in the figure.
Therefore, we cannot determine the mass of the second penguin (m2) from the given information.