Two toy ducks attached to each other by a string, are being pulled by a child. The front duck is m1= 5.0 kg and the back duck is m2= 2.0 kg. The happy guy pulls them with a force of 10 N and friction exists. The toys are moving with constant velocity.
A) if the coefficient of kinetic friction is = 0.12, find the force of friction which is acting on the second duck.
B) calculate the tension in the string connecting the ducks
A) Ah, the ol' ducks-on-a-string problem. To find the force of friction on the second duck, we'll need to first determine the overall force acting on the ducks. Since they're moving at a constant velocity, we know the net force must be zero. So, the force of friction must equal the force applied by the happy guy.
The force applied by the happy guy is 10 N, so the force of friction on the second duck must also be 10 N. Friction always opposes motion, so it acts in the opposite direction of the applied force.
B) As for the tension in the string connecting the ducks, it's equal throughout the entire string. So, the tension is the same at both ends.
Since there's no acceleration, we know the net force in the horizontal direction must be zero. The only horizontal forces acting on the system are the applied force by the guy and the force of friction. Therefore, the tension in the string must also equal the applied force, 10 N.
So, the tension in the string connecting the ducks is 10 N. Just think of it like the ducks are in a tug-of-war with the happy guy!
To solve this problem, we need to consider the forces acting on the ducks. The forces acting on the ducks are the applied force (10 N) and the force of friction.
A) The force of friction can be calculated using the equation:
Force of friction = coefficient of friction * normal force
The normal force acting on an object can be calculated using the equation:
Normal force = mass * acceleration due to gravity
For the front duck (m1 = 5.0 kg):
Normal force1 = m1 * g = 5.0 kg * 9.8 m/s^2 = 49 N
For the back duck (m2 = 2.0 kg):
Normal force2 = m2 * g = 2.0 kg * 9.8 m/s^2 = 19.6 N
The force of friction1 acting on the front duck can be calculated as:
Force of friction1 = coefficient of friction * normal force1 = 0.12 * 49 N = 5.88 N
The force of friction2 acting on the second duck can be calculated as:
Force of friction2 = coefficient of friction * normal force2 = 0.12 * 19.6 N = 2.352 N
Therefore, the force of friction acting on the second duck is 2.352 N.
B) To calculate the tension in the string connecting the ducks, we need to consider the net force acting on the ducks. Since the ducks are moving with constant velocity, the net force acting on them is zero.
The applied force can be considered as the force exerted by the child on the front duck. This force is transmitted through the string between the ducks.
The tension in the string is equal to the force exerted by the front duck on the second duck, which is also equal to the force exerted by the second duck on the front duck.
Therefore, the tension in the string is 10 N.
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). In this case, since the ducks are moving with constant velocity, we know that the net force acting on them is zero.
A) To find the force of friction acting on the second duck, we need to consider the forces acting on it. There are three main forces at play: the force of tension in the string, the force of friction acting opposite to the direction of motion, and the force of the child pulling the ducks.
Since the ducks are moving with constant velocity, the force of tension in the string (T) must be equal in magnitude and opposite in direction to the force of friction acting on the back duck. Therefore, to calculate the force of friction on the second duck, we can simply find the tension in the string.
B) To calculate the tension in the string connecting the ducks, we need to consider the forces acting on both ducks. The forces on the front duck are the force of tension in the string (T) pulling it forward and the force of the child pulling it backward. The forces on the back duck are the force of tension in the string (T) pulling it forward and the force of friction acting opposite to the direction of motion.
Given:
m1 = 5.0 kg (mass of the front duck)
m2 = 2.0 kg (mass of the back duck)
F = 10 N (force applied by the child)
µ = 0.12 (coefficient of kinetic friction)
Using the information above, let's solve the problem step by step.
A) The force of friction on the second duck is equal to the tension in the string:
T = force of friction on the second duck
Since the net force on the ducks is zero, we can set up the following equation:
ΣF = m2a = 0
The only horizontal forces acting on the second duck are the force of tension (T) and the force of friction (f). The force of tension is in the positive direction, and the force of friction is in the negative direction. Therefore, we can express this equation as:
T - f = m2a
Since the ducks are moving with constant velocity, their acceleration (a) is zero, and the equation becomes:
T - f = 0
We can rearrange this equation to solve for the force of friction:
f = T
So, the force of friction on the second duck is equal to the tension in the string.
B) To calculate the tension in the string, let's consider the forces acting on both ducks:
For the front duck (m1):
The forces acting on the front duck are the force of tension (T) pulling it forward and the force of the child pulling it backward:
ΣF = T - F = m1a = 0
Rearranging the equation, we find:
T - F = 0
T = F
So, the tension in the string connecting the ducks is equal to the force applied by the child.
Therefore, the force of friction acting on the second duck is equal to the force of tension in the string (10 N), and the tension in the string connecting the ducks is also 10 N.