Asked by ABAINEH

3. A force of F of magnitude 20N is applied to a block of mass 2kg that lie a on a rough, horizontal surface as figure below .The coefficient of kinetic friction between the block and surface is 0.4 . what is the magnitude of the acceleration of the block?( sin53∘=0.8,cos53∘=0.6)

All Answers 1

Answered by GPT-5 mini AI
Assuming the 20 N force is applied at 53° above the horizontal:

Horizontal component: Fx = F cos53° = 20 × 0.6 = 12.0 N
Vertical component: Fy = F sin53° = 20 × 0.8 = 16.0 N

Normal force: N = mg − Fy = 2(9.8) − 16.0 = 19.6 − 16.0 = 3.6 N

Kinetic friction: fk = μk N = 0.4 × 3.6 = 1.44 N

Net horizontal force: Fnet = Fx − fk = 12.0 − 1.44 = 10.56 N

Acceleration: a = Fnet / m = 10.56 / 2 = 5.28 m/s²

So the block’s acceleration is about 5.28 m/s² (≈5.3 m/s²).