(a).A 220-kg sleigh is pulled by one horse at constant velocity for 0.90 km on a snowy, horizontal surface, the coefficient of kinetic friction between the sleigh runners and snow is 0.27. What power must the horse deliver to the sleigh for the trip to take 9 minutes?

(b) Image a motor to lift heavy objects.What must be the minimum power of the motor to lift 750-kg load at 1.0 m/s?

Power=work/time= force*distance/time

= mg*mu*900/timein secnds

power= mg*velocity

(a) To solve for the power the horse must deliver to the sleigh, we can use the equation:

Power = Force × Velocity

First, let's determine the force experienced by the sleigh due to friction. The force of kinetic friction can be found using the equation:

Force of Kinetic Friction = coefficient of kinetic friction × Normal Force

The normal force is the force exerted by the surface perpendicular to the object's weight. Since the sleigh is on a horizontal surface, the normal force is equal to the weight of the sleigh.

Normal Force = Weight of the Sleigh = mass × gravitational acceleration

Weight of the Sleigh = 220 kg × 9.8 m/s^2 = 2156 N

Force of Kinetic Friction = 0.27 × 2156 N = 583.32 N

Since the sleigh is at constant velocity, its net force must be zero. Therefore, the force exerted by the horse pulling the sleigh is equal in magnitude but opposite in direction to the force of kinetic friction.

Force exerted by the Horse = Force of Kinetic Friction = 583.32 N

Now, convert the distance traveled from kilometers to meters:
0.90 km = 0.90 km × 1000 m/km = 900 m

The work done by the horse is given by the equation:

Work = Force × Distance

The horse must do work against the force of kinetic friction, which is equal to the force exerted by the horse.

Work = Force of Kinetic Friction × Distance = 583.32 N × 900 m = 524,988 J

To determine the power, which is the work done per unit time, divide the work by the time taken:

Power = Work / Time

Convert the time from minutes to seconds:
9 minutes = 9 minutes × 60 s/minute = 540 s

Power = 524,988 J / 540 s ≈ 972.19 W

Therefore, the horse must deliver approximately 972.19 Watts of power to the sleigh for the trip to take 9 minutes.

(b) To calculate the minimum power required for a motor to lift a 750-kg load at 1.0 m/s, we can use the equation:

Power = Force × Velocity

The force required to lift the load against gravity can be calculated using the equation:

Force = mass × gravitational acceleration

Force = 750 kg × 9.8 m/s^2 = 7350 N

The velocity is given as 1.0 m/s.

Now, substitute the values into the power equation:

Power = Force × Velocity = 7350 N × 1.0 m/s

Power = 7350 Watts

Therefore, the minimum power of the motor required to lift a 750-kg load at 1.0 m/s is 7350 Watts.