What is the Time to Reach the Top of an Inclined Plane with a Free-Moving Box?

[weesiang_loke]

Homework Statement

Consider a small box mass m initially at the bottom of an inclined plane mass M, length L with angle of inclination of \theta. The surface between the plane and the block and the plane and the horizontal are both frictionless. A force F is applied horizontally to the small box. Need to find the time when the small box reached the top of the inclined plane.

Homework Equations

The Attempt at a Solution

I have:
F - Nsin\theta = ma_m,x
Ncos\theta = ma_m,y
Nsin\theta = Ma_M,x

for the relative motion of the small box to the inclined plane:
tan\theta =\frac{<b>a</b>_m,y}{<b>a</b>_m,x - <b>a</b>_M,x}

then i try to use the distance traveled in y direction, so
1/2 * a_m,y * t^2 = L sin \theta

I am not sure they are the correct equations.

the answer given is t = \sqrt{\frac{2<b>L</b>(1+(<b>m</b>/<b>M</b>)(sin\theta)^2)}{(<b>F</b>/<b>m</b>)cos\theta - g(1+<b>m</b>/<b>M</b>)sin\theta}}Thanks for any help given :-)

 

[weesiang_loke]

well i have solved it. thanks for anyone who tried or trying to do this question.

cheers,
weesiang_loke