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Question 4 A car of mass 820 kg has a maximum power of 30 kW and moves against a constant resistance of motion to 910 N. Calculate the maximum speed of the car in the following situation; b)Up an incline (hill) of 8.75° to the horizontal?​

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Answer:

P = W / t = m g s / t = m g v       where work by auto = m g s

30 kw = 30000 watts = 30000 J / s

Work wasted = F v       as shown above relating work and power

Work done against incline = m g s sin 8.75  and power against incline

= m g v sin 8.75 = 1222 v Joules / sec

power in moving auto = power available - power lost to friction

power in moving auto = 30000 - resistance = 30000 - 910 v

1222 v = 30000 - 910 v

v = 30000 / 2132 = 14 m/s

Note: constant resistance to motion must mean P = W / t = F s / t = F v

Maximum speed will be v = 14.07 m/s

What is power ?

Power is defined time rate of doing work or delivering energy.

Work done = Force * displacement

Power = work done / time

           = (Force * displacement ) / time

since , displacement / time = velocity

Power  = force * (displacement / time ) = force  * velocity

Power = mass * acceleration * velocity (  as force = mass * acceleration)

since , car is travelling up in an incline of 8.75° to the horizontal

hence , it will have two components of velocity

vertical component = v sin (theta ) = v sin(8.75°)

horizontal component = v cos (theta)= v cos (8.75°)

Power = m* a * v sin (theta )

          = 820 * 9.8  * v *sin (8.75°)                        ( a = g = 9.8 m[tex]/s^{2}[/tex])

         = 1221.472 v

Power lost due to resistance in motion  = 910 * velocity = 910 v

Power = Total power - power lost due to resistance in motion

           = 30 * [tex]10^{3}[/tex]  - 910 v

     

30 * [tex]10^{3}[/tex]  - 910 v = 1221.472 v  

v = 14.07 m/s

Maximum speed will be v = 14.07 m/s

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