Question

As in Problem A above, a block of mass M starts from rest and is pushed up a frictionless ramp inclined at an angle θ above the horizontal by a constant force F. Use the work-energy theorem to find the block’s speed after it has been pushed a distance L up the incline. Your answer should be written in terms of M, θ, F, L and the acceleration due to gravity, g.

Answer 1

Answer:

A(i)

The solution to this question is shown on the second uploaded image

A(ii)

The final speed is $v = \sqrt{2L (\frac{F}{M} - gsin\theta )}$

B

The block speed after a distance L  is  $v= \sqrt{2L (\frac{F}{M} -gsin \theta )}$

Explanation:

From the question  

The net force i the x-direction is mathematically represented as

               $F_{net} = Ma$

From the the diagram in the second uploaded image we see that

             $F_{net} = F - Mgsin \theta$

Therefore

              $F- Mgsin\theta = Ma$

Making a the subject

              $a = \frac{F}{M} - gsin\theta$

Applying the law of motion

                $v^2 =u^2 + 2as$

where  u = 0 m/s and s =L

       $v^2 = 0 + 2(\frac{F}{M} - gsin \theta )L$

=>   $v = \sqrt{2L (\frac{F}{M} - gsin\theta )}$

According to Energy conservation law and work theorem

           Workdone by F + Workdone by gravity = change in kinetic energy

Mathematically this is given as

                $F * L - (mgsin \theta)L = \frac{1}{2} M (v^2-u^2)$

   Since u = 0 m/s

                    $L (\frac{F}{M} - gsin \theta ) = \frac{1}{2} v^2$

                  $v= \sqrt{2L (\frac{F}{M} -gsin \theta )}$