A block of mass M is pushed by a force F up a smooth plane that is inclined at an angle q with respect to the horizontal as shown below. The next three questions pertain to this situation. The symbol g has the value +9.8 m/s2.

In order to accelerate the block up the plane, F must be greater than Mg.

I think this statement is true but in what way I can explain it?

If M = 2.0 kg, F = 24 N, and q = 30o, what is t, the time it takes the block to move a distance d = 5 m up the plane if the block were initially at rest?

t = 0.83 s
t = 0.91 s
t = 1.19 s
t = 1.43 s
t = 1.55 s


https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/hwb/05/01/quiz5_2.gif

Show in what direction the weight on the block acts. You will find that the weight acts in a vertical direction.

So, break this force into two components, namely a component along the plane and another perpendicular to the plane.

Then find the instance when the resultant force along the plane is zero. As soon as force F overcomes the component along the plane, there will be a resultant force and block will move up.

For the second part, find the resultant force.

Use F = ma to find the acceleration of the block.

Then, use:

d = \frac12 at^2

to find the time.

with the given idea, I have formulated that a=(cos30)*24/2, and then put a into the equation, however, I can't get the choices the question is given

I have this similar problem, just with different number, so I try to work this one out, but cannot get any of the answer posted

You missed something.

Along the plane, the force up the plane is 24 N

Down the plane, the force that pulls back the block is given by mg sin(30) = (2*9.8*0.5) = 9.8 N

Net force along the plane = 24 - 9.8 = 14.2 N

Then, you apply F = ma

Remember that in the equation F = ma, F is the resultant force.

Just a quick thought; if the force is applied in an horizontal way (parallel to the floor) to the block, will there be anything that is different?

Yes, then there will be no vertical component of the force and the equation will then become:

N = mg

You need to know that the reaction force onto the block is dependent on the forces acting on the mass.

Remember Newton's Third law. The mutual forces of action and reaction between two bodies are equal, opposite and collinear.

So, if the forces pushing the block down increase, the normal reaction increases, when the forces pushing the block down decrease, the normal reaction decreases.

EDIT: This is what happens when the block is on horizontal ground.

is the value 14.2N normal force?

So by this explanation, the new normal force from the ramp on the block compare to the original one would be the same?

Hm.. You asked me if the block was on the horizontal floor, what would happen right?

If that was so, then the normal reaction on the block when a horizontal force is applied on it will equal mg.

If the same force, with same direction acts on the block when it is on the ramp, then, you have another situation.

When you break up the force into components, you get one along the plane and another one perpendicular to the plane. Resolving perpendicular to the plane, we get:

Fsin\theta + mg = N

14.2 N is the net force along the plane, not the force perpendicular to the plane, the direction in which the normal reaction acts.

And could you please use the answer box instead of the comments button please? Thanks.

the ansewer is 1.19s. 24/2-9.8sin(30)=7.1=a
5=7.1t^2/2
t=1.19s.
Does any one know if it is true or false for f must be greater than mg

In this question, F > mgsin\theta

The component of gravity that acts along the plane is mgsin\theta and this is a fraction of mg.

So, can be equal to mg, but still, the block will move upwards.

Let's give numbers. If \theta = 30

Then, the component along the plane becomes mgsin(30) = \frac12 mg

In this particular case, F must be greater than 0.5mg. F could be equal to 0.6mg and the block would still go up.