A 57 kg boy and a 45 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface.
If the acceleration of the girl toward the boy is 1.5 m/s2, determine the magnitude of the acceleration of the boy toward the girl.
Answer in units of m/s2
Since the same force is acting on both,
the acceleration of the boy will be less
by their mass ratio:
a=(45kg/57kg) * 1.5m/s^2 = 1.184 m/s^2.
Alternate Method.
Fg = ma = 45 * 1.5 = 67.5 N. = Force
acting on girl.
Fb = Fg = 67.5 N.
a = Fb/m = 67.5 / 57 = 1.184 m/s^2. =
Acceleration of boy.
Well, well, well, looks like we have a classic case of a tug-of-war on ice! It's always slippery when you're pulling on elastic ropes, isn't it?
Now, let's tackle this problem head-on. We have a 57 kg boy and a 45 kg girl. The girl is accelerating towards the boy at a rate of 1.5 m/s2. We need to find the magnitude of the boy's acceleration towards the girl.
Now, because this is a frictionless icy surface and we're dealing with an elastic rope, we can assume that the tension in the rope is the same for both the boy and the girl. And we all know that the force equals mass times acceleration, right?
So, the force on the girl is given by F = m * a, where m is the girl's mass (45 kg) and a is her acceleration (1.5 m/s2). That gives us a force of 45 kg * 1.5 m/s2 = 67.5 N.
Since the tension in the rope is the same for the boy, the force on the boy is also 67.5 N.
Now, let's use Newton's second law, which tells us that force equals mass times acceleration. So, we have:
67.5 N = m * a_boy
Substituting the boy's mass (57 kg) for m, we can now solve for a_boy:
67.5 N = 57 kg * a_boy
Dividing both sides by 57 kg, we get:
1.18 m/s2 = a_boy
So, the magnitude of the acceleration of the boy towards the girl is approximately 1.18 m/s2.
Just remember, when it comes to tug-of-war on ice, it's all about slipping and sliding with a touch of physics!
To determine the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion:
F = m * a
where F is the net force, m is the mass, and a is the acceleration.
Since the boy and the girl are connected by the elastic rope, the net force acting on them is equal in magnitude and opposite in direction (according to Newton's third law).
Let's assume the acceleration of the boy toward the girl is a2. The net force acting on the girl can be calculated as:
Fg = mg * a
where Fg is the net force on the girl, mg is the mass of the girl, and a is the given acceleration (1.5 m/s^2).
Similarly, the net force acting on the boy can be calculated as:
Fb = mb * a2
where Fb is the net force on the boy, mb is the mass of the boy, and a2 is the acceleration we need to find.
Since the net forces on the boy and the girl are equal in magnitude and opposite in direction, we have:
Fg = Fb
mg * a = mb * a2
From the given information, we have:
mg = 45 kg
mb = 57 kg
a = 1.5 m/s^2
Plugging in these values, we can solve for a2:
45 kg * 1.5 m/s^2 = 57 kg * a2
67.5 kg m/s^2 = 57 kg * a2
Dividing both sides of the equation by 57 kg, we get:
1.18 m/s^2 = a2
Therefore, the magnitude of the acceleration of the boy toward the girl is 1.18 m/s^2.
To determine the magnitude of the acceleration of the boy toward the girl, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's denote the acceleration of the boy toward the girl as "a" (in m/s^2).
According to the problem, the girl has a mass of 45 kg, and her acceleration is 1.5 m/s^2. Therefore, the force acting on the girl can be calculated using Newton's second law:
Force on the girl = mass of the girl × acceleration of the girl
Force on the girl = 45 kg × 1.5 m/s^2
Force on the girl = 67.5 N
Now, since the boy and the girl are engaged in a tug-of-war and connected by an elastic rope, the force acting on the boy will be equal in magnitude but opposite in direction to the force acting on the girl due to Newton's third law of motion (every action has an equal and opposite reaction).
Therefore, the force acting on the boy is also 67.5 N.
Next, we can determine the mass of the boy, which is given as 57 kg.
Finally, we can use Newton's second law to calculate the acceleration of the boy toward the girl:
Force on the boy = mass of the boy × acceleration of the boy
67.5 N = 57 kg × acceleration of the boy
Solving for acceleration of the boy, we get:
acceleration of the boy = 67.5 N / 57 kg
acceleration of the boy ≈ 1.1842 m/s^2
Therefore, the magnitude of the acceleration of the boy toward the girl is approximately 1.1842 m/s^2.