Two blocks are arranged. M1 is on the table and m2 is hanging from the pulley. The pulley can be considered to be massless, and friction is negligible. M1 is four times more massive than M2.

If the system is released from rest, how far will M1 travel in 0.963 s?

I'll bet that your instructor will be happy to see the 12 physics homework questions you've posted here. <g>

To determine how far M1 will travel in 0.963 seconds, we can start by analyzing the forces acting on the system.

Since the pulley is assumed to be massless and there is negligible friction, we can assume that the tension in the rope is the same on both sides of the pulley.

Let's denote the mass of M2 as m2 and the mass of M1 as m1. Given that m1 is four times more massive than m2, we can write the equation: m1 = 4m2.

Since the system is released from rest, we can assume it starts from rest and hence the initial velocity of M1 is 0 m/s. We can also assume that the acceleration of the system is constant.

Using Newton's second law, we can write the equation for the net force on M1 as: F1 = m1 * a, where F1 is the net force acting on M1 in the horizontal direction, m1 is the mass of M1, and a is the acceleration of the system.

The net force acting on M1 is equal to the tension in the rope (T) minus the gravitational force pulling M1 downward (m1 * g), where g is the acceleration due to gravity.

Equating the forces, we have: T - m1 * g = m1 * a.

Since the tension in the rope is the same on both sides of the pulley, we can write another equation for the net force acting on M2: T - m2 * g = m2 * a.

From these two equations, we can substitute T from the second equation into the first equation, giving us: (m2 * g + m1 * g) - m1 * g = m1 * a.

Simplifying, we have: m2 * g = m1 * a.

Since g is a constant, we can cancel it out, leaving us with: m2 = m1 * a.

Substituting the given relationship between m1 and m2 (m1 = 4m2), we now have: m2 = 4m2 * a.

Simplifying, we have: 1 = 4a.

Therefore, a = 1/4. This means the acceleration of the system is 1/4 m/s^2.

To determine how far M1 will travel in 0.963 seconds, we can use the equation: x = x0 + v0 * t + 0.5 * a * t^2.

Since M1 starts from rest, the initial position (x0) is 0, and the initial velocity (v0) is also 0.

Plugging in the values, we have: x = 0 + 0 * 0.963 + 0.5 * (1/4) * (0.963)^2.

Simplifying further, we have: x = 0 + 0 + 0.5 * (1/4) * (0.927369)^2.

Evaluating the expression, we find: x ≈ 0.0602 meters.

Therefore, M1 will travel approximately 0.0602 meters in 0.963 seconds.