Two blocks are connected by a massless rope that passes over a pulley. M1= 2 kg and M2= 4 kg. The pulley is 12cm in diameter and has a mass of 2kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.5 Nm. if the blocks are released from rest, how long does it take the 4 kg block to reach the floor from a height of 1 m?

Well, it sounds like we have a classic physics problem on our hands! But don't worry, I'll do my best to clown around and simplify it for you.

First, we need to find the acceleration of the system. Since the two blocks are connected by a rope passing over a pulley, they will have the same acceleration. Let's call this acceleration "a".

To find "a", we can use the following equation: ΣF = (M1 + M2) * a, where ΣF is the sum of the forces acting on the system.

The forces acting on the system are the gravitational forces pulling the blocks down. For M1, we have a force of M1 * g (mass of M1 times the acceleration due to gravity). And for M2, the force is M2 * g.

Now, let's plug in the given values. We have M1 = 2 kg, M2 = 4 kg, and g ≈ 9.8 m/s².

For M1: F1 = M1 * g = 2 kg * 9.8 m/s² = 19.6 N

For M2: F2 = M2 * g = 4 kg * 9.8 m/s² = 39.2 N

So, the sum of the forces is ΣF = F2 - F1 = 39.2 N - 19.6 N = 19.6 N.

Now, let's plug this value into the equation:

19.6 N = (2 kg + 4 kg) * a

Simplifying:

19.6 N = 6 kg * a

a ≈ 3.27 m/s²

Now that we have the acceleration, let's figure out the time it takes for the 4 kg block to reach the floor from a height of 1 m.

We can use the following equation: Δd = V0 * t + 0.5 * a * t².

Since the 4 kg block is released from rest, its initial velocity (V0) is 0. And Δd is the distance it travels, which is 1 m.

1 m = 0 * t + 0.5 * 3.27 m/s² * t²

Rearranging the equation:

0.5 * 3.27 m/s² * t² = 1 m

Solving for t:

t² ≈ (1 m) / (0.5 * 3.27 m/s²) ≈ 0.306 s²

t ≈ √(0.306 s²) ≈ 0.553 s

So, it takes approximately 0.553 seconds for the 4 kg block to reach the floor from a height of 1 m.

I hope my clowning around made this problem a little less daunting for you!

To solve this problem, we can use the principle of conservation of mechanical energy. The potential energy of the 4 kg block at a height of 1 m is equal to the sum of its kinetic energy and the potential energy of the 2 kg block at the same height.

Let's start by calculating the potential energy of the 4 kg block:

Potential Energy (PE) = mass * gravity * height
PE1 = 4 kg * 9.8 m/s^2 * 1 m
PE1 = 39.2 J

Next, we can calculate the potential energy of the 2 kg block using the same formula:

PE2 = 2 kg * 9.8 m/s^2 * 1 m
PE2 = 19.6 J

Now, we know that the total potential energy is conserved, so:

PE1 + PE2 = KE1 + KE2

Since both blocks are initially at rest, their initial kinetic energy is zero, so:

PE1 + PE2 = KE1 + KE2
39.2 J + 19.6 J = 0 + KE2

Simplifying this equation:

59.8 J = KE2

We also know that the total mechanical energy is conserved, so the sum of the kinetic energy and the potential energy of the system remains constant throughout the motion. Therefore, when the 4 kg block reaches the floor, its potential energy will be zero, and all the initial potential energy will be converted into its kinetic energy.

KE2 = 59.8 J

Using the formula for kinetic energy:

KE = (1/2) * mass * velocity^2

We can solve for the velocity of the 4 kg block when it reaches the floor:

59.8 J = (1/2) * 4 kg * velocity^2
119.6 J = 4 kg * velocity^2
29.9 m/s^2 = velocity^2
velocity = sqrt(29.9 m/s^2)
velocity = 5.47 m/s

Now that we know the velocity of the 4 kg block when it reaches the floor, we can calculate the time it takes to reach the floor using the following equation:

v = u + at

Here, u is the initial velocity (which is 0 since the block is released from rest), a is the acceleration, and t is the time taken.

Using the formula:

5.47 m/s = 0 + 9.8 m/s^2 * t
5.47 m/s = 9.8 m/s^2 * t
t = 5.47 m/s / 9.8 m/s^2
t = 0.557 s

Therefore, it takes approximately 0.557 seconds for the 4 kg block to reach the floor from a height of 1 m.

To solve this problem, we need to determine the acceleration of the system and then use the kinematic equation to find the time it takes for the 4 kg block to reach the floor.

First, let's determine the net force acting on the system. The force due to gravity on the 2 kg block is given by F1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, F1 = 2 kg * 9.8 m/s^2 = 19.6 N. The force due to gravity on the 4 kg block is F2 = m2 * g, where m2 is the mass of the block. Therefore, F2 = 4 kg * 9.8 m/s^2 = 39.2 N.

Next, let's determine the tension in the rope. Since the rope is massless, the tension is the same on both sides of the pulley. Let T be the tension in the rope. The net force acting on the 2 kg block is F1 - T, and the net force acting on the 4 kg block is T - F2. Since the blocks are connected by a rope, their acceleration is the same. Therefore,

F1 - T = m1 * a,
T - F2 = m2 * a,

where a is the acceleration of the system.

Solving these two equations simultaneously, we get:

19.6 N - T = 2 kg * a,
T - 39.2 N = 4 kg * a.

Rearranging these equations gives:

T = 19.6 N - 2 kg * a,
T = 4 kg * a + 39.2 N.

Setting the two expressions for T equal to each other:

19.6 N - 2 kg * a = 4 kg * a + 39.2 N.

Simplifying and solving for a gives:

6 kg * a = 19.6 N - 39.2 N,
6 kg * a = -19.6 N.

Dividing by 6 kg, we find:

a = -19.6 N / 6 kg,
a = -3.27 m/s^2.

The negative sign indicates that the acceleration is in the opposite direction to the initial displacement (i.e., the blocks are going down).

Now, using the kinematic equation:

h = (1/2) * a * t^2,

where h is the height from which the 4 kg block is dropped, a is the acceleration, and t is the time it takes for the block to reach the floor.

Plugging in the known values, we get:

1 m = (1/2) * (-3.27 m/s^2) * t^2.

Simplifying, we find:

2 = -1.635 m/s^2 * t^2,
t^2 = -2 / -1.635,
t^2 ≈ 1.22,
t ≈ √(1.22),
t ≈ 1.10 s.

Therefore, it takes approximately 1.10 seconds for the 4 kg block to reach the floor from a height of 1 m.