Question

A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D along the direction of its motion.
Part B
By what multiplicative factor RK does the initial kinetic energy increase, and by what multiplicative factor RWdoes the work done by the force increase (with respect to the case when the particle had a mass M)?
If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.

You should enter the two factors separated by a comma.

Part C
The particle's change in speed over the distance D will be ______ the change in speed when it had a mass equal to M.

Part D
By what factor RK does the initial kinetic energy increase (with respect to the first situation, with mass M and speed vi), and by what factor RW does the work done by the force increase?
Again, enter the two factors, separated by a comma.

Part E
The particle's change in speed over the distance D will be ______ the change in speed when it had an initial velocity equal to vi.

Answer 1

Answer:

Explanation:

PART A

From Newtons second law:

$F=ma\\a=\frac{F}{m}\\\frac{dv}{dt}=\frac{F}{m}\\\\(\frac{dv}{ds}\times\frac{ds}{dt})=\frac{F}{m}\\\\v\frac{dv}{ds}=\frac{F}{m}\\vdv=\frac{F}{m}ds$

Integrating above expression by applying limits:

$\int\limits^{vf}_{v_{i}} {vdv} =\frac{F}{m}\int\limits^5_0ds$

Here Distance =D

$\frac{v^2_f-v^2_i}{2}=\frac{FD}{m}$

The final speed of the particle after travelling distance D

$v_f=\sqrt{v^2_i+\frac{2FD}{m}}$

PART B

The Kinetic energy of the particle of mass M is,

$K_1=\frac{1}{2}Mv^2$

For $M=3M\\K_2=\frac{1}{2}(3M)v^2=3(K_1)$

The kinetic energy increases by factor 3

The workdone depends on the factor and displacement of the body. Therefore, workdone increases by factor 1.

PART C

Final velocity is,

$v_f=\sqrt{v^2_i+\frac{2FD}{m}}$

Here,$\frac{F}{m}$ represents the acceleration $a$

For$m=3M\\\\a=\frac{F}{2M}$

In this case acceleration is decreased so, velocity also decreases.

PART D

Initial kinetic energy is  

$K_1=\frac{1}{2}Mv^2$

Here $v=3v\\\\K_1=\frac{1}{2}M(2v)^2\\\\=[tex]K_1=9(\frac{1}{2}M(v)^2)$

Kinetic energy increases by factor of 9. workdone remains same even as velocity increases. Therefore, workdone increases by factor 1

PART E

The final speed of the particle is

$v^2_{f'}=v^2_i+\frac{2FD}{m}$....(1)

The final speed of the particle with speed $3v_i$ is,

$v^2_{f'}=9v^2_i+\frac{2FD}{m}$....(2)

substract two equations:

$v_{f'}-v_f=\frac{8v_i^2}{(v_{f'}+v_f)}\\\\=\frac{8v_i}{\sqrt{9}+\sqrt{1}}\\=2v_i$

Thus the particle changes in speed over distance D and will be less with an initial speed equal to $v_i$