Asked by jindi
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 6.9 N, acting 54° north of west. What is the magnitude of the body's acceleration?
Answers
Answered by
drwls
Add the two force vectors.
The components are:
East: 9.0 - 6.9 cos54 = 4.944 N
North: 6.0 sin 54 = 4.864 N
Then calculate the magnitude of the resultant.
sqrt[(4.944)^2 + (4.864)^2] = ___
Divide that by the mass to get the acceleration, in m/s^2.
The components are:
East: 9.0 - 6.9 cos54 = 4.944 N
North: 6.0 sin 54 = 4.864 N
Then calculate the magnitude of the resultant.
sqrt[(4.944)^2 + (4.864)^2] = ___
Divide that by the mass to get the acceleration, in m/s^2.
Answered by
rhiannon
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