Question

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?

Answer 1

total mass M= 280 kg

The coefficient of kinetic friction between the sled and the snow  = 0.080

The sled acceleration a =0.65 m/s2

Initial speed u = 0

cruising speed v = 15 km/h.

= 15 (1000m/3600s) = 4.166 m/s

(A ). The team's maximum power output during the acceleration phase P = ?

Net force Ma = Applied force - fricitonal force

= F - Mg

From this F = Ma +Mg

= M(a+g)

= 280(0.65+(0.08x9.8))

= 401.52 N

So, P = Fv

= 401.52 x 4.166

= 1,672.7323 W

 

Answer 2

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.

$v = 15 \frac{1000m}{3600s} = 4.166m/s$

Part (A ).

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

Net force = Ma = Applied force - fricitonal force

$Ma = F - \mu Mg$

From this

$F = Ma +\muMg$

$F = 280(0.65+(0.08x9.8))$

$F = 401.52 N$

Now we have P = Fv

$P = 401.52 \times 4.166 = 1,672.7 W$

Part (B)

Now after reaching the cruising speed

acceleration = 0

$F - \mu mg = 0$

$F = \mu m g = 0.08 \times 280 \times 9.8$

$F = 219.52 N$

now power at cruising speed is given as

$P = F.v$

$P = 219.52 \times 4.166 = 914.5 W$