Physics help?

Question

Got a tough one here.
A 2.kg block on a frictionless plane inclined at 30 degrees from horizontal has a force F acting on it upward parallel to the incline. The block is accerlating downward at 2.0 m/s. Find F.

I assumed it was the F=ma eqation

F = 2kg*2.0 = 4N

I know its not that easy and cos 30 comes in somewhere.. can you point me in the right direction?

Answer 1

There is another very easy way to solve this problem using dynamics. First, draw a free body diagram and take the sum of the forces in the x and y components.

The sum of the forces in the y component is n - mgcos30 = 0. Therefore n = mgcos30...however we won't even need this information to solve the problem since all the motion is in the x component.

The sum of the forces in the x component is F - mgsin30 = ma.
we can rearrange this to get an expression for the force.

F = ma + mgsin30

F = m(a+gsin30)

F = (2.0 kg)[ (-2.0 m/s/s) + (9.8 m/s/s)(sin30)]

sin30 = 1/2, so F = (2.0 kg)[ (-2.0 m/s/s) + (4.9 m/s/s)]

F = (2.0 kg)(2.9 m/s/s)

F = 5.8 N

Answer 2

You know that the 2kg block has gravity pulling it toward the earth with a force of (2kg)*g, where g is the acceleration due to gravity. Draw a diagram. You'll see that the force of gravity is 60 degrees away from the inclined plane. That is, you can draw a triangle with one leg sitting on the inclined plane, another leg perpendicular to the inclined plane, and the force of gravity connecting the two. The angle between the hypotenuse (the force of gravity) and the leg on the inclined plane is 60 degrees. That means that the force that is being directed down the hill is

(2kg)*g*cos(60 degrees)

I use a cos(60 degrees) because cos(60 degrees)=adjacent/hypotenuse, thus adjacent=hypotenuse*cos(60 degrees), and the hypotenuse is the force of gravity, which is (2kg)*g. Note that the force perpendicular to the hill is the force which will be canceled by the normal force. Since the plane is frictionless, I'm not going to discuss the normal force.

Now you know the force down the inclined plane. On top of that, you know that there is a force acting up the inclined plane of F. You know that:

(2kg)*g*cos(60 degrees) - F = (2kg)*(2 m/s/s)

That is, assuming that "downward" is the positive direction, then (2kg)*g*cos(60 degrees) is a positive force and F is a negative force. If you sum the forces, you get the net force. That net force is what is causing the 2 m/s/s acceleration. The magnitude of the net force is (2kg)*(2 m/s/s), which is 4 Newtons. Now we have enough to solve for F:

F = (2kg)*g*cos(60 degrees) - (2kg)*(2 m/s/s)
= (2kg)*g*cos(60 degrees) - 4N
= (2kg)*g*(0.5) - 4N
= (1kg)*g - 4N

Now, if you assume that g=9.81m/s/s, then you get:

F = 9.81N - 4N = 5.81N

Thus, the force pulling the block up the hill is 5.81N. Since it is less than the force pulling it down the hill (which is due to gravity), then the block accelerates down the hill at 2.0 m/s/s.

Answer 3

Yeah something like that.