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) A.10m/s2 B. 5.2m/s2 C. 4m/s2
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Assume the force is applied 53° above the horizontal. Components:
- Horizontal: Fx = 20 cos53° = 20(0.6) = 12 N
- Vertical up: Fy = 20 sin53° = 20(0.8) = 16 N
Take g = 10 m/s² (common in such problems). Normal force:
N = mg − Fy = 2·10 − 16 = 20 − 16 = 4 N
Kinetic friction:
fk = μk N = 0.4·4 = 1.6 N
Net horizontal force:
Fnet = Fx − fk = 12 − 1.6 = 10.4 N
Acceleration:
a = Fnet / m = 10.4 / 2 = 5.2 m/s²
Answer: B (5.2 m/s²). (If g = 9.8 m/s², a ≈ 5.28 m/s².)
- Horizontal: Fx = 20 cos53° = 20(0.6) = 12 N
- Vertical up: Fy = 20 sin53° = 20(0.8) = 16 N
Take g = 10 m/s² (common in such problems). Normal force:
N = mg − Fy = 2·10 − 16 = 20 − 16 = 4 N
Kinetic friction:
fk = μk N = 0.4·4 = 1.6 N
Net horizontal force:
Fnet = Fx − fk = 12 − 1.6 = 10.4 N
Acceleration:
a = Fnet / m = 10.4 / 2 = 5.2 m/s²
Answer: B (5.2 m/s²). (If g = 9.8 m/s², a ≈ 5.28 m/s².)
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