A 110-kg man lowers himself to the ground from a height of 12 m by holding on to a rope passed

A 110-kg man lowers himself to the ground from a height of...

Key Concepts

Coupled Motion Dynamics

When two objects are connected by a rope over a pulley, their motions are interdependent. The acceleration of one mass will directly affect the other because any movement of one mass results in a corresponding movement of the other due to the rope constraint. This dynamic is important in solving the problem since the man’s descent and the sandbag's rise both contribute to the energy conversion process and the resultant motion of the system.

Controlled Descent Methods

This concept refers to strategies that can reduce the final speed during a descent. By actively applying forces, such as increasing the tension in the rope or using friction (for example, by braking the rope with hand forces or a mechanical device), one can convert kinetic energy into other forms, like heat, thereby reducing the acceleration and the final impact speed. This underscores that the use of controlled energy dissipation can safely lower the speed at which one reaches the ground.

Conservation of Mechanical Energy

This concept involves the idea that in the absence of non-conservative forces like friction, the total mechanical energy (the sum of potential and kinetic energy) remains constant. In systems where objects change height, gravitational potential energy is converted into kinetic energy or vice versa. In the given problem, the change in the man’s potential energy as he descends is used to determine the kinetic energy and thus his final speed when he reaches the ground.

Pulley System Mechanics

A frictionless pulley changes the direction of the force applied by a rope without changing its magnitude, allowing the conversion of forces between connected masses. This concept is crucial in understanding how the dynamics of one body (the man) are interrelated with another (the sandbag). The arrangement means that energy and tensions are shared across the rope, and the inertial differences between the masses influence the overall acceleration and energy distribution in the system.

Transcript

00:01 Hello student we are going to understand this question here given mass of man let it is represented by capital m which is given over 110 k sly hide from the ground that it is represented by small m which is given 12 meter and mass of sandbag that it is represented by small m which is given 7 24 .0 kg.

00:40 So we have to find in bit a we have to find with what speed does the man hits the ground? that the speed by which man is hitting the ground man is getting the ground the ground is v so let's draw the diagram here with the help of this diagram we can see that this sandbag is moving in upward direction with acceleration a and mind is moving in the downward direction with acceleration a and tension in the string is t so with the help of this blog diagram we can draw the equation tension is in upper direction and weight of the sandbag is in downward direction and net acceleration is in upper direction so we can write it as t minus m g is equal to m into a let this is equation number one and for man that will be capital m into g minus t is equal to capital m in 2a that this is equation number 2 now from equation 1 and 2 when we will add equation 1 and 2 then we will get m g minus small m g is equal to m a plus capital now here we can take common g capital m x x capital m minus small m is equal to a into small m plus capital m.

02:53 Now we can write a is equal to z m minus small m upon m plus small m plus capital m.

03:06 Now substituting the value in this, a is equal to value of g.

03:19 G is gravitational acceleration that is 9 .8 meter per second square.

03:24 And value of capital m means value of man, mass of man that is 110 and mass of sandbag that is 74 upon small m value of small m that is 74 plus mass of man that is 110 now further solving a is equal to 9.

03:55 8 into 36 upon 1304 after solving we will get 1 .91 meter per second square this will be the resultant acceleration now let the final velocity by which it is getting head getting hit to the ground we have supposed that is big now applying newton's third law, that is v square is equal to g square plus 2a into h.

04:40 Here final velocity is v.

04:42 Initial velocity is 0.

04:45 2 into value of a, we have calculated 1 .91...

Please give Ace some feedback