Six dogs pull a two-person sled with a total mass of 280 kg. the coefficient of kinetic friction between the sled and the snow is 0.080. the sled accelerates at 0.65 m/s2 until it reaches a cruising speed of 15 km/h. what is the team's maximum power output during the acceleration phase? what is the team's power output during the cruising phase?

The power output during the acceleration phase is 1.69 KW. The power output during the cursing phase is 0.932 KW.

We have to convert the velocity to m/s as follows;

15 km/h × 1000/3600 = 4.16 m/s

We know that;

Net force = Applied force - frictional force

Net force = ma = 280 kg × 0.65 m/s2 = 182 N

Applied force = ?

Frictional force = μmg =  0.080 × 280 kg  × 10m/s

Frictional force = 224 N

Applied force = Net force + frictional force

Applied force = 182 N +  224 N

Applied force = 406 N

Power output during acceleration phase = 406 N × 4.16 m/s = 1.69 KW

During the cruising phase, a = 0

Hence, Net force = 0

Applied force = Frictional force = 224 N

Power output = 224 N × 4.16 m/s = 0.932 KW

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As it is given that

total mass M= 280 kg

coefficient of kinetic friction = 0.080

acceleration a =0.65 m/s2

Initial speed = 0

cruising speed v = 15 km/h.

Part (A ).

For the team's maximum power output during the acceleration phase

Net force = Ma = Applied force - fricitonal force

From this

Now we have P = Fv

Part (B)

Now after reaching the cruising speed

acceleration = 0

now power at cruising speed is given as