Answer:
A(i)
The solution to this question is shown on the second uploaded image
A(ii)
The final speed is [tex]v = \sqrt{2L (\frac{F}{M} - gsin\theta )}[/tex]
B
The block speed after a distance L is [tex]v= \sqrt{2L (\frac{F}{M} -gsin \theta )}[/tex]
Explanation:
From the question
The net force i the x-direction is mathematically represented as
[tex]F_{net} = Ma[/tex]
From the the diagram in the second uploaded image we see that
[tex]F_{net} = F - Mgsin \theta[/tex]
Therefore
[tex]F- Mgsin\theta = Ma[/tex]
Making a the subject
[tex]a = \frac{F}{M} - gsin\theta[/tex]
Applying the law of motion
[tex]v^2 =u^2 + 2as[/tex]
where u = 0 m/s and s =L
[tex]v^2 = 0 + 2(\frac{F}{M} - gsin \theta )L[/tex]
=> [tex]v = \sqrt{2L (\frac{F}{M} - gsin\theta )}[/tex]
According to Energy conservation law and work theorem
Workdone by F + Workdone by gravity = change in kinetic energy
Mathematically this is given as
[tex]F * L - (mgsin \theta)L = \frac{1}{2} M (v^2-u^2)[/tex]
Since u = 0 m/s
[tex]L (\frac{F}{M} - gsin \theta ) = \frac{1}{2} v^2[/tex]
[tex]v= \sqrt{2L (\frac{F}{M} -gsin \theta )}[/tex]
