Asked by aly
A 7.6 kg chair is pushed across a frictionless floor with a force of 42 N that is applied at an angle of 22° downward from the horizontal. What is the magnitude of the acceleration of the chair?
Answers
Answered by
Ann Hiro
F = MA, right? Just mix in a cos Θ to calculate effective force (42N x cos 22° = 38.94172N), and solve.
38.94172 = 7.6A
A = 38.94172 / 7.6
A = 5.123911m/s^2
38.94172 = 7.6A
A = 38.94172 / 7.6
A = 5.123911m/s^2
Answered by
drwls
a = F_x/m
F_x is the x-component of F, 42 cos22 = 38.94 N
AnnHiro is correct. Thanks for helping us out, Ann.
The answer should only be quoted to two significant figures, 5.1 m/s^2. The input numbers have comparable accuracy.
F_x is the x-component of F, 42 cos22 = 38.94 N
AnnHiro is correct. Thanks for helping us out, Ann.
The answer should only be quoted to two significant figures, 5.1 m/s^2. The input numbers have comparable accuracy.
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