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A 50 kg crate rests on a horizontal ground for which the coefficient of friction is k =0.3. A towing force of 400 N is applied to the crate at an angle of 300 to the horizontal. Find the velocity of the crate in 5 s starting from rest.
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\[a= (F _{1}-F _{2})/M\] \[F _{applied}=400N\] \[F _{f}=\mu F _{N}\] \[F _{f}=(.3)(50)(9.8)\] a=5.06m/s^2 v=at v=25.3m/s
so, we don't have to use the given angle ?
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