Posted by Alex on Sunday, December 5, 2010 at 8:03am.
It is not clear to me
1. how a block will "rest" on a smooth inclined plane, and
2. in which direction is the horizontal force acting, towards up the plane, or down?
3. The first paragraph seems incomplete
I don't know.....it's written like that in the book...... It's being pushed up....... the answer is 12.9m/s.....But I dun know how to calculate it......
Do you have a diagram that shows the angle of the incline and the direction of the force?
no....sorry I forgot to mention, the angle is 20 degree.....
Resolve forces along the inclined plane of inclination θ=20°:
For the mass, m = 0.5 kg
and the applied force, F = 16N
component of force down the incline,
Fm = mg*sin(θ)
Due to force F, component up the incline,
Ff = Fcos(θ)
Net force up,
Fup
= Ff-Fm
= Fcos(θ)-mgsin(θ)
Acceleration up the plane
a = F/m
Initially at rest, u=0, and final velocity v after travelling a distance of S = 5m,
use the relation
v²-u² = 2aS
Solve for v in m/s. I get 12.9 m/s.
a= F/m
=16/0.5
=32m/s
v²=2*32*5
v =17.9
But I get 17.9m/s..... acceleration is equal to net force over mass or just force over mass?
Anything wrong with my steps?
Can I minus 17.9 by 5N which is the friction?
I get it now......Thanks alot.....