A block of mass M is pulled along a smooth horizontal floor by a force of magnitude F inclined at an angle q as shown in the figure below. The next two questions pertain to this situation. The symbol g has the value +9.8 m/s2.

Which of the following statements concerning N, the vertical force that the floor exerts on the block, is true?

N < Mg
N = Mg
N > Mg

If M = 1.5 kg, F = 6.0 N, and q = 45o, what is a, the magnitude of the acceleration of the block?

a = 2.0 m/s2
a = 2.8 m/s2
a = 4.0 m/s2
a = 5.7 m/s2
a = 6.9 m/s2

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/hwb/05/01/quiz5_1.gif

1.

Simply try to apply Newton's first law.

You don't expect the block to dig into the floor or rise from the floor, so, the net vertical force on the block must be zero.

What are the forces acting in a vertical plane to the box?

From this, tell us what you get for this question.

2.

Find the resultant force along the horizontal.

Use F = ma

And find a :)

So, for the first one, if you said that the block should stay in the plane, so N = mg because they cancel out each other and the block will not neither fly off nor dig into the floor, right?

So, for the first one, if you said that the block should stay in the plane, so N = mg because they cancel out each other and the block will not neither fly off nor dig into the floor, right?

You didn't take into consideration the force F applied. F has an effect on the block.

Resolving F along the vertical gives Fsin\theta

So, total forces on block is 0.

Hence:

Fsin\theta + N = mg

Do you see it now? :)