At t = 1.0 s, a 0.40-kg object is falling with a speed of...

Name: Problem 20, Chapter 6: Pearson Physics
Uploaded: 2023-02-18T01:21:24-08:00
Duration: 4 min 7 s
Channel: Prabhakar Kumar
Description: At t = 1.0 s, a 0.40-kg object is falling with a speed of 6.0 m>s. At t = 2.0 s, it has a kinetic energy of 25 J. (a) What is the kinetic energy of the object at t = 1.0 s? (b) What is the speed of the object at t = 2.0 s? (c) How much work was done on the object between t = 1.0 s and t = 2.0 s?

At t = 1.0 s, a 0.40-kg object is falling with a speed of 6.0 m>s. At t = 2.0 s, it has a kinetic energy of 25 J. (a) What is the kinetic energy of the object at t = 1.0 s? (b) What is the speed of the object at t = 2.0 s? (c) How much work was done on the object between t = 1.0 s and t = 2.0 s?

At t = 1.0 s, a 0.40-kg object is falling with a speed of 6.0 m>s.
At t = 2.0 s, it has a kinetic energy of 25 J.
(a) What is the kinetic energy of the object at t = 1.0 s?
(b) What is the speed of the object at t = 2.0 s?
(c) How much work was done on the object between t = 1.0 s and
t = 2.0 s?

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = ½mv², where m is the mass and v is the velocity of the object. It reflects how changes in velocity or mass can alter the energy of an object in motion.

Work-Energy Principle

The work-energy principle states that the net work done on an object is equal to its change in kinetic energy. This concept links force and displacement (work) with the subsequent energy changes experienced by the object, allowing one to determine how forces acting over distances affect speed and energy.

Transcript

00:01 In this video we are going to focus on the given information where we have the information about an object which is having the mass 0 .40 kgs and then we can say that it is given that it is falling with a speed of 6 meter per second.

00:15 So it is given over here that the kinetic energy is here equals to 25 joules at a certain instant and that very instant is basically here 2 second.

00:28 Now, in the first part of this very problem, we are asked what will be the value of kinetic energy of the object at the instant of t is equal to one second and we have to find it in terms of joules.

00:42 So we can say that kinetic energy will be here equals to half m and then it will be here v square.

00:51 So we have the velocity, right? and what is the value of velocity? let us write over here and that is here equals to.

00:58 6 meter per second.

01:00 So this very value is basically given at t is equal to 1 second.

01:05 So we can use this very value over here in the first part of this very problem right.

01:09 So it can be here written as half multiplied with mass which is 0 .40 multiplied with 6 raise to the power 2.

01:17 So the value of kinetic energy is here going to come around.

01:21 Let us write its value and that will be basically 7 .2 joules and then we can here focus on finding out the answer for the second part of this very question so in the second part it can be here written that we are required to find out the value of speed of the object at a given instant and that very instant is t is equals to two seconds right so we have the value that kinetic energy at two second is equal to 25 or basically we can here say that it will be half and then here it is mass and then here it will be v square so this v value is the velocity at t is equal to two second do not get confused with the value that we have written in the above part so it is here velocity at t is equals to two second it will be better if we write it suppose v1 so it is our v1 square and thus we can say that if we rearrange the equation then it can be here written that v1 is is equal to under root of it will be here two times kinetic energy divided with m and if we plug in the value then it will be here two times the kinetic energy is here equals to we can say 25 and in the denominator we are having the value of mass which is equal to 0 .40 and hence it is here going to come around 11 .180 and then it will be here meter per second so we have find out the value of of velocity here as 11 .180 and this can be put inside a box in order to highlight it.

03:00 So let us now see how we can find out the answer for the third part of this very question...

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