Physics 12: Power Output Calculation for Accelerating Object

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To calculate the power output of a 5.0 kg object accelerated from rest to 6.0 m/s over 2.0 m with a friction force of 4.0 N, the correct approach involves using kinematics to determine the time taken for acceleration. The formula d = vt is not suitable here due to the object's acceleration, necessitating the use of kinematic equations. After finding the time, power can be calculated using P = (F_net * d) / t, where F_net is the net force after accounting for friction. The discussion highlights the importance of using appropriate equations for varying motion conditions. Accurate calculations yield a power output of 270 Watts.
gdhillon
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A 5.0 kg object is accelerated from rest to 6.0 m/s while moving 2.0 m across a level
surface. If the force of friction is 4.0 N, calculate the power output



I used d=vt to find the time which was .33sec then i put that into P=Fd/t and had (49-4)(2)/.35 and got 270 Watts
 
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Cool - was there a question in all that?
Is there some role the v-t graph could play?
 
Last edited:
d = vt is applicable only for constant velocities. The object is accelerating in this case. You must use a kinematics equation to find the time - you know the initial and final velocities, and the distance traveled.
 

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