Two blocks are arranged. The pulley can be considered to be massless, and friction is negligible. M1 is four times more massive than M2. (m1 is on the table and m2 is hanging from the pulley).
If the system is released from rest, how far will M1 travel in 0.963 s?
Well, M1 seems to be quite the traveler! But you want to know how far it will travel in 0.963 seconds? That's a specific time frame! Where is M1 going? Is it off on a vacation? Does it have a Tinder date to get to? Or maybe it just wants to impress M2 with its incredible velocity! One can only wonder... But of course, I'm here to help with some math too!
To calculate how far M1 will travel in 0.963 seconds, we need to know the acceleration of the system. So, do you have any more information about that, or are we just left to speculate about M1's adventures?
To determine how far M1 will travel in 0.963 s, we need to consider the acceleration of the system.
Given that M1 is four times more massive than M2, let's denote the mass of M2 as m and the mass of M1 as 4m.
We can start by finding the net force acting on the system. Since the pulley is massless and friction is negligible, the only force acting on the system is the force due to the weight of M2, which can be calculated as m2g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
The net force can be expressed as:
Net force = m2g
According to Newton's second law, net force is equal to the mass multiplied by acceleration. In this case, the acceleration of the system is the same for both M1 and M2.
Net force = (m2 + 4m) * a
Now we can set these two equations equal to each other:
m2g = (m2 + 4m) * a
To find the acceleration of the system, we need to use the rotational dynamics. The acceleration of a point on the edge of a rotating object, like the pulley, is given by:
a = α * r
Where α is the angular acceleration and r is the radius of the pulley.
Since the pulley is released from rest, its angular velocity is initially zero, and α is related to the linear acceleration a by the equation:
α = a / r
Substituting this into the previous equation, we get:
m2g = (m2 + 4m) * (a / r)
Now, we can solve for a:
a = (m2g * r) / (m2 + 4m)
Next, we can use the equation for linear displacement (s) in terms of initial velocity (v0), time (t), and acceleration (a):
s = v0 * t + (1/2) * a * t²
Since the system is released from rest, the initial velocity is zero:
s = (1/2) * a * t²
Plugging in the values for a and t, we can calculate the distance M1 will travel:
s = (1/2) * ((m2g * r) / (m2 + 4m)) * t²
Now you can plug in the specific values for m2, g, r, m, and t to find the distance M1 will travel in 0.963 s.