Question

In: Physics

A truck accelerates uniformly from rest to 19.5 m/s in 5.3 s along a level stretch...

A truck accelerates uniformly from rest to 19.5 m/s in 5.3 s along a level stretch of road. Determine the average power required to accelerate the truck for the following values of the weight (ignore friction).

(a) 1.00x10^4 N

(b) 1.20 x10^4 N

Solutions

Expert Solution

Gravitational acceleration = g = 9.81 m/s2

Part (a)

Weight of the truck = W = 1 x 10^4 N

Mass of the truck = m

m = 1019 kg

Initial speed of the truck = V1 = 0 m/s

Final speed of the truck = V2 = 19.5 m/s

Time taken to reach 19.5 m/s = T = 5.3 sec

Average power required to accelerate the truck = P

The work done is equal to the change in kinetic energy of the truck.

P = 3.66 x 10^4 W

Part (b)

Weight of the truck = W = 1.2 x 10^4 N

Mass of the truck = m

m = 1223 kg

Initial speed of the truck = V1 = 0 m/s

Final speed of the truck = V2 = 19.5 m/s

Time taken to reach 19.5 m/s = T = 5.3 sec

Average power required to accelerate the truck = P

The work done is equal to the change in kinetic energy of the truck.

P = 4.39 x 10^4 W

a) Average power required to accelerate a truck weighing 1x10^4 N = 3.66 x 10^4 W

b) Average power required to accelerate a truck weighing 1.2x10^4 N = 4.39 x 10^4 W

genius_generous answered 2 months ago

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