1) FGround = -Fweight = mg = 1200 kg x 9.8 N/kg iin the opposite direction of weight (so up)
2) Ff (only force in horizontal direction) = ma = 1200kg x 1.5 m/s2 in the direction of acceleration.
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Nayan N.
asked 06/28/222) The same car accelerates out of the stoplight at 1.5 m/s2. find the force of friction (magnitude and direction) that the road puts on the wheels.
1) FGround = -Fweight = mg = 1200 kg x 9.8 N/kg iin the opposite direction of weight (so up)
2) Ff (only force in horizontal direction) = ma = 1200kg x 1.5 m/s2 in the direction of acceleration.
Please consider a tutor. Take care.
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