Force F on Mass m at Angle Θ: Acceleration Calculations

[dreamz25]

q) A force F acts on a block of mass m placed on a horizontal smooth surface at an angle Θ with horizontal. Then

(A) If F sinΘ < mg then a = (F + mg) / m

(B) Acceleration = FcosΘ/m where, F > mg cosecΘ

(C) Acceleration = F/m if FsinΘ > mg

(D) If F SinΘ > mg then a = (F + mg) / m

my work
--------

on drawing the F.B.D we we get..

clearly there will be linear and vertical acceleration as well...
Now if F sinΘ > mg then the Normal force will be 0
=> F sinΘ - mg = ma_y ---- (i)

and, F cosΘ = ma_x ------ (ii)

Adding the two eqns. we get,

F sinΘ + F cosΘ - mg = m (a_x + a_y)

or, a_x + a_y = (F - mg) / m

but the correct option is d.

couldn't represent vectors with their notations so please understand
it urselves.

Am i wrong? if yes then where and why?

Thanks in advance...!

 

[Doc Al]

None of those answers look right. Are you sure there's not a typo in there? (Check the orientation of the > signs.)

And you can't just add perpendicular vector components, such as a_x + a_y.

 

[dreamz25]

why not...?

 

[Doc Al]

why not...?

They are vectors. Direction counts.

 

[asteriatic]

Your work is correct. The correct option is (D) because when F sinΘ > mg, the normal force will be greater than mg, resulting in a positive acceleration in the x-direction. This means that the net force in the x-direction is F + mg, and dividing by the mass gives the correct expression for acceleration. Option (A) is incorrect because it does not take into account the normal force. Option (B) is incorrect because it assumes that F is greater than mg, which may not always be the case. Option (C) is incorrect because it assumes that the normal force is equal to mg, which is not always true. Overall, your explanation and calculations are correct.