p05 001

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1. We apply Newton’s second law (specifically, Eq. 5-2).

(a) We find the x component of the force is

F

x

= ma

x

= ma cos 20

=(1.00 kg)(2.00 m/s

2

) cos 20

= 1.88 N .

(b) The y component of the force is

F

y

= ma

y

= ma sin 20

=(1.0 kg)(2.00 m/s

2

) sin 20

= 0.684 N .

(c) In unit-vector notation, the force vector (in Newtons) is



F = F

x

ˆi+ F

y

ˆj = 1.88ˆi+ 0.684ˆj .


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