This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 47. The free-body diagram (not to scale) for the block is shown below. FN is the normal force exerted by the floor and mg is the force of gravity. (a) The x component of Newton’s second law is F cosθ = ma, where m is the mass of the block and a is the x component of its acceleration. We obtain a= 12.0 N cos 25.0° F cosθ = = 2.18 m / s2 . 5.00 kg m b g This is its acceleration provided it remains in contact with the floor. Assuming it does, we find the value of FN (and if FN is positive, then the assumption is true but if FN is negative then the block leaves the floor). The y component of Newton’s second law becomes so FN + F sinθ – mg = 0, FN = mg – F sinθ = (5.00 kg)(9.80 m/s2) – (12.0 N) sin 25.0° = 43.9 N. Hence the block remains on the floor and its acceleration is a = 2.18 m/s2. (b) If F is the minimum force for which the block leaves the floor, then FN = 0 and the y component of the acceleration vanishes. The y component of the second law becomes F sinθ – mg = 0
2 mg ( 5.00 kg ) ( 9.80 m/s ) F= = = 116N. sin θ sin 25.0° (c) The acceleration is still in the x direction and is still given by the equation developed in part (a): F cos θ (116 N) cos 25.0° a= = = 21.0m/s 2 . m 5.00 kg ...
View Full Document