A 14.0 kg box is pulled along a horizontal frictionless surface by a force of 33.0 N at a 35.5 ∘ angle with respect to the surface.

Question

A 14.0 kg box is pulled along a horizontal frictionless surface by a force of 33.0 N at a 35.5 ∘ angle with respect to the surface. What is the horizontal acceleration of the box?

Express your answer using three significant figures. Calculate the normal force on the box.

Express your answer using three significant figures.

Answer 1

Hello Al C,

If we draw a free body diagram with all forces properly resolved , we get following equations

33 Sin theta -14g = N....... Eq 1 Where theta = 35.5

where N = Normal reaction ( acting perpendicular to the surface) and

33 Cos theta = 14 a..... Eq2 , where a = Acceleration of the block ( Parallel to the surface

Plug in

Sin theta = Sin (35.5) = 0.581

and Cos theta= Cos (35.5) = 0.814

you get N = Normal force

and a= Acceleration of block