In Fig. E6.7 assume that there is no friction force on the...

In Fig. E6.7 assume that there is no friction force on the 20.0 -N block that sits on the tabletop. The pulley is light and frictionless. (a) Calculate the tension $T$ in the light string that connects the blocks. (b) For a displacement in which the $12.0-\mathrm{N}$ block descends $1.20 \mathrm{m},$ calculate the total work done on (i) the $20.0 .0-\mathrm{N}$ block and (ii) the 12.0 -N block. (c) For the displacement in part (b), calculate the total work done on the system of the two blocks. How does your answer compare to the work done on the 12.0 -N block by gravity? (d) If the system is released from rest, what is the speed of the 12.0 -N block when it has descended 1.20 $\mathrm{m}$ ?

Key Concepts

Newton's Second Law of Motion

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, expressed as F = ma. It is fundamental in determining the acceleration of objects in a system when forces such as tension and gravity are applied, allowing for the analysis of motion in connected bodies.

Frictionless Systems

A frictionless system is one in which no resistive forces oppose the motion of objects. This simplifies the analysis by ensuring that the only forces doing work are those like gravity and tension, and makes the conservation of mechanical energy directly applicable between kinetic and potential energies.

Uniform Tension in a Light String

In a system with a light, inextensible string and a frictionless pulley, the tension is the same throughout the string. This uniformity allows the forces transmitted through the string to be analyzed consistently, linking the accelerations of different masses and simplifying the force equations in the system.

Work Done by a Force

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force, typically calculated as W = Fd cos?. This concept is used to quantify the energy transferred to or from an object when it is moved by a force.

Work-Energy Theorem

This theorem relates the net work done on an object to its change in kinetic energy. It provides a powerful method for solving dynamics problems by linking forces and displacements directly to the motion characteristics of a system without needing to calculate acceleration explicitly.

Conservation of Mechanical Energy

In the absence of non-conservative forces such as friction, the total mechanical energy (potential plus kinetic energy) of a system remains constant. This conservation principle is used to equate the work done by conservative forces, like gravity, to changes in kinetic energy, aiding in the determination of speeds and displacements.

Transcript

00:01 High in the given problem there is a table top or which a block of 20 newton has been kept it is a pulley a frictionless must less pulley in the edge of this table over which a light string is passing and another block has been suspended with it whose weight is 12 newton its weight will be acting vertically downward which will create a tension t in the string in both the segments of the string the tension is this now as the weight of the first block is w1 is equal to 20 newton so if i compare it with m1g and we can take acceleration due to gravity as 10 meter per second square so mass of the first block comes out to be 2kg similarly for the weight of the second block which is 12 newton it can be written as m2g or m2 into 10 so its mass will come out to be 1 .2 kg now using the free body diagram of this hanging block the net force acting on it is 12 minus t which is given by m1 sorry this is m2a using newton's second lock motion or we can say this is 1 .2a we make it equation number one now using the free body diagram of this lock put on the table top we get t is equal to m1a or we can say two eyes of a so putting this value of tension here in this equation number one we get 12 minus 2a is equal to 1 .2 a or we can say 12 is equal to 3 .2 a so acceleration of the system comes out to be 12 by 3 .2 meter per second square or it is 3 .75 meter per second square now for the first part of the problem we put the value of this acceleration in equation number two so doing that we get tension t is equal to twice of a or twice of 3 .75 so this tension comes out to be 7 .5 newton this is the answer for the first part of the problem now for the second part we have to find work done on both the blocks if they move a distance of 1 .20 meter so for the first part you have to find work done on 20 newton block so simply it will become the force which is nothing but equal to tension into the displacement so it will become 7 .5 which is the tension 7 .5 newton into displacement 1 .2 so this work will come out to be 9 june.

03:43 In the second part we have to find the work done on 12 newton block.

03:49 So here on the hanging block the net force is 12 minus t so this is 12 minus t is the net force and the displacement will be same as we know the tension is 7 .5 newtons so it becomes 12 minus 7 .7 .5 newton so it becomes 12 minus 7.

04:09 5 into 1 .2 so the work comes out to be 5 .4 jules so these are the two answers of the second part now for the third part we have to find the total work done on both the blocks it will be equal to algebraic sum of w20 and w12 so it comes out to be 9 plus 5 .4 joules which becomes 14 .4 jules which becomes 14 .4 and in the same part we have to find work done by the gravity on 12 newton block so it will be equal to force applied by the gravity on the block which is nothing but the weight of the block it of the hanging block 12 newton into distance travel which will be same 1 .2 meter so this work comes out to be 14 .4 joules so we conclude that the total work done on the blocks is nothing but equal to work done by the gravity on hanging block...

Please give Ace some feedback