Starting at time t=0, net force F1 is applied to an object...

Starting at time $t=0$, net force $F_{1}$ is applied to an object that is initially at rest. (a) If the force remains constant with magnitude $F_{1}$ while the object moves a distance $d$, the final speed of the object is $v_{1} .$ What is the final speed $v_{2}$ (in terms of $v_{1}$ ) if the net force is $F_{2}=2 F_{1}$ and the object moves the same distance $d$ while the force is being applied? (b) If the force $F_{1}$ remains constant while it is applied for a time $T,$ the final speed of the object is $v_{1} .$ What is the final speed $v_{2}$ (in terms of $v_{1}$ ) if the applied force is $F_{2}=2 F_{1}$ and is constant while it is applied for the same time $T ?$ In a later chapter we'll call force times distance work and force times time impulse and associate work and impulse with the change in speed.)

Key Concepts

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). This principle underpins both the work-energy and impulse-momentum theorems by describing how forces affect an object's motion, whether the force is applied over a distance or over a period of time.

Work-Energy Theorem

This theorem relates the work done on an object by a force to the change in its kinetic energy. When a constant force is applied over a displacement, the product of the force and the distance (work) increases the object's kinetic energy, which in turn determines its final speed. It is fundamental in understanding how different forces applied over the same distance result in different speeds.

Impulse-Momentum Theorem

This theorem connects the impulse imparted to an object (the product of force and the time interval over which it is applied) to the change in its momentum. When a constant force is applied for a given time, the resulting impulse produces a change in speed. This concept explains how variations in the applied force, even if applied over the same time interval, lead to different momentum changes and thus different final speeds.

Transcript

00:01 Hi, in the given problem, net force acting on the object at a time t is equal to 0 is f1.

00:15 Then initially speed of the object is 0.

00:21 In first part of the problem, distance covered by the object is d.

00:29 Final speed achieved by the object is given as v.

00:37 So using newton's second law of motion, which says f is equal to force equals to the product of mass and acceleration.

00:48 Acceleration in the motion of the object comes out to be f by m as the force here is f1, so it becomes f1 by m.

00:59 Now using third equation of motion which says vf square is equal to v i square plus 2 as for vf this is v1 square is equal to 0 as the initial speed was 0 plus 2 for acceleration this is f1 by m and distance traveled was d.

01:27 So using that v1 comes out to be square root of 2 f1 d by m.

01:41 Now then the force is doubled.

01:46 So f2 is made twice to f1.

01:50 In that case, acceleration in the motion of the object a2 will be given by f2 by m.

01:56 So it will become twice of f1 by m.

02:01 So now using again the third equation of motion v2 square is equal to for vi i again this is 0 plus two times of a and a is this time to f1 by m into d.

02:18 So this v2 will come out to be 2 twos of 4 f1 by m into d.

02:28 So it is clear that as here this was v1 square was equal to 2 f1 d by m.

02:38 So 2 times of 2 f1 d by m can be used as v1 and this is square...

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