<< What is the magnitude of block's acceleration? >>
I will ASSUME that the "horizontal force of magnitude 40 N" is directed parallel to the incline. Thus being said,
50 - F = ma
where
F = frictional force acting on the block
m = mass of the block
a = acceleration of the block
Since
F = µN = µ(mgcos 37), then the above equation becomes
50 - µ(mgcos 37) = ma
and substituting appropriate values,
50 - 0.310(5.10)(9.8)(cos 37) = 5.1(a)
and solving for "a"
a = 7.38 m/sec^2
<< The block's initial speed is 4.10 m/s. How far up the plane does the block go? >>
Use the law of conservation of energy,
KE @ bottom of incline - Friction loss = PE when block stops
KE @ bottom of incline = (1/2)(5.1)(4.10)^2 = 42.86 joules
Friction loss = 0.310(5.1)(9.8)(sin 37)(L)
PE when block stops = 5.1(9.8)(L*sin 37)
where
L = distance along incline travelled by block
The above equation then becomes,
42.86 - 0.310(5.1)(9.8)(sin 37)(L) = 5.1(9.8)(L*sin 37)
and solving for "L"
L = 1.09 m
Hope this helps.
