A 0.500 kg block, attached to a spring with length 0.60 m and force constant 40.0 N / m, is a

A 0.500 kg block, attached to a spring with length 0.60 m...

A $0.500 \mathrm{~kg}$ block, attached to a spring with length $0.60 \mathrm{~m}$ and force constant $40.0 \mathrm{~N} / \mathrm{m},$ is at rest with the back of the block at point $A$ on a frictionless, horizontal air table (Fig. $\mathbf{P 7 . 6 9}$ ). The mass of the spring is negligible. You move the block to the right along the surface by pulling with a constant $20.0 \mathrm{~N}$ horizontal force. (a) What is the block's speed when the back of the block reaches point $B,$ which is $0.25 \mathrm{~m}$ to the right of point $A ?$ (b) When the back of the block reaches point $B$, you let go of the block. In the subsequent motion, how close does the block get to the wall where the left end of the spring is attached?

Key Concepts

Work-Energy Principle

This principle states that the work done on an object by external forces results in a change in the object's kinetic energy. In problems like this, where a constant force moves a block over a distance, calculating the work (force multiplied by displacement) is essential for understanding how the block gains speed.

Elastic Potential Energy

Elastic potential energy is the energy stored in a spring when it is either compressed or stretched away from its equilibrium position. It is characterized by a quadratic relationship to the displacement (x) of the spring from equilibrium and is calculated using the formula ½ k x². This energy becomes available when the spring returns to its equilibrium state, converting into kinetic energy.

Conservation of Mechanical Energy

The conservation of mechanical energy in frictionless systems implies that the total energy—comprising kinetic energy and potential energy (including elastic potential energy)—remains constant. When the block and the spring interact after the force is released, the energy exchanges between kinetic and elastic potential forms, allowing one to predict motion characteristics such as maximum compression or speed.

Hooke's Law

Hooke's Law describes how the force exerted by a spring is directly proportional to the displacement from its equilibrium position, with the constant of proportionality being the spring constant, k. This law is crucial in understanding how the spring resists changes in its shape and how force is transmitted between the block and the spring.

Transcript

00:01 High in the given problem, mass of the block which is attached to the horizontal spring is given as 0 .50 kilogram.

00:18 Original length of the spring is l is equal to 0 .6 meter and spring factor or force constant of the spring is given as k is equal to 40 .0 newton per meter.

00:50 In the first part of the problem, a force of 20 .0 newton is applied on this block which is attached with the spring towards right under which this block moves through a distance d given by 0 .25 meter.

01:24 So work done on the block by the supplied force will be consumed in two parts.

01:31 One of its part will be gained by the block as its kinetic energy and another will be stored in the spring in this stretched spring as its elastic potential energy.

01:42 So using work energy theorem, work done on the block by the force applied will be equal to kinetic energy of the block plus elastic potential energy of the stretched spring.

02:29 Hence using the expression for the work done that will be f into d is equal to kinetic energy of the block half mv square plus elastic potential energy of the spring half k d square.

02:49 So this half mv square may be given as fb minus half k d square plugging in all known values here.

03:05 For f, this is 20, newton into 0 .25 meter minus half into 40 into square of 0 .25.

03:17 And here this is half into mass of the block 0 .5 into v square.

03:28 So here it becomes 5 minus 1 .25.

03:42 It comes 3 .5.

03:43 It comes out to be 3.

03:44 Hence, v square will become 2 times of 3 .75 divided by 0 .5, which comes out to be 15.

03:56 Hence, the speed of the block at this moment will be square root of 15 meter per second, or we can say this is v, is equal to 3 .87 meter per second, which becomes the answer for the first part of the problem...

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