A BLOCK OF MASS M=5kg IS RESTING ON A ROUGH HORIZONTAL SURFACE FOR WHICH THE CO EFFICIENT OF FRICTION IS 0.2 WHEN FORCE F=40N IS APPLIED THE ACCELERATION OF THE BLOCK WILL BE (g=10m/s)
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56
mass of block = 5kg
F = 40N
μ = 0.2
g = 10 m/s²
friction, f = μmg = 0.2×5×10 = 10N
net force =F-f = 40-10 = 30N
acceleration = 30/5 = 6m/s²
F = 40N
μ = 0.2
g = 10 m/s²
friction, f = μmg = 0.2×5×10 = 10N
net force =F-f = 40-10 = 30N
acceleration = 30/5 = 6m/s²
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4
Answer:
6 m/s^2
Explanation:
hope it will help you
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