Asked by Alex

A block of mass 0.5 kg rests on a smooth inclined plane. It is then pushed by a horizontal force of 16N. When the block has displaced 5m along the inclined plane.

What would be the speed of the block if the plane is rough and a friction of 5N is acting on the block?
Answered by MathMate
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
Answered by Alex
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......
Answered by MathMate
Do you have a diagram that shows the angle of the incline and the direction of the force?
Answered by Alex
no....sorry I forgot to mention, the angle is 20 degree.....
Answered by MathMate
Resolve forces <i>along</i> the inclined plane of inclination θ=20°:

For the mass, m = 0.5 kg
and the applied force, F = 16N
component of force <i>down</i> the incline,
Fm = mg*sin(θ)

Due to force F, component <i>up</i> 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.

Answered by Pedro
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?
Answered by Alex
I get it now......Thanks alot.....
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