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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?

Respuesta :

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.

[tex]v = 15 \frac{1000m}{3600s} = 4.166m/s[/tex]

Part (A ).

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

Net force = Ma = Applied force - fricitonal force

[tex]Ma = F - \mu Mg[/tex]

From this

[tex]F = Ma +\muMg[/tex]

[tex]F = 280(0.65+(0.08x9.8))[/tex]

[tex]F = 401.52 N[/tex]

Now we have P = Fv

[tex]P = 401.52 \times 4.166 = 1,672.7 W[/tex]

Part (B)

Now after reaching the cruising speed

acceleration = 0

[tex]F - \mu mg = 0[/tex]

[tex]F = \mu m g = 0.08 \times 280 \times 9.8[/tex]

[tex]F = 219.52 N[/tex]

now power at cruising speed is given as

[tex]P = F.v[/tex]

[tex]P = 219.52 \times 4.166 = 914.5 W[/tex]

pstnonsonjoku

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

Learn more: https://brainly.com/question/13155407

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