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=When the mass m = 5kg is placed on a smooth horizontal plane, what is the acceleration of the object when the horizontal force F 1 = 6N acts on the object When an object with a mass of 5kg is placed on a smooth horizontal plane, what is the acceleration of the object when the horizontal force F1 = 6N acts on the object? If another horizontal force F = 8N is applied and F1 and F2 are in the same horizontal plane, F1 is perpendicular to F2. What is the acceleration of the object?
First question: a = f / M = 6 / 5 = 1.2
Second question: according to the principle of vector synthesis of force, it is easy to know that the resultant force is 10N, so a = f / M = 10 / 5 = 2
Please answer:
What is the acceleration of an object with a mass of 0.3KG under 6N force
A=F/m=6/0.3=20N/m
The common point force (F1) = (LG2, LG2), (F2) = (lg5, LG2) acts on the object and produces displacement (s) = (2lg5,1)
Then the work W of the common point force on the object is
() denotes a vector
Resultant force F = (LG2, LG2) + (lg5, LG2) = (LG2 + lg5, LG2 + LG2) = (1,2lg2)
So w = F.S = (1,2lg2). (2lg5 + 1) = 1 * 2lg5 + 2lg2 * 1 = 2 (lg5 + LG2) = 2
Where. Is vector dot multiplication and * is general multiplication
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