Question

A sleigh and driver with a total mass of 130 kg are pulled up a hill with a 15∘ incline by a horse, as illustrated in the figure(Figure 1).

Part A

Part complete

If the overall retarding frictional force is 850 N and the sled moves up the hill with a constant velocity of 5.0 km/h , what is the power output of the horse? (Express in horsepower, of course.)

Express your answer using two significant figures.

P =

hp

Part B

Suppose that in a spurt of energy, the horse accelerates the sled uniformly from 5.0 km/h to 18 km/h in 7.0 s . What is the horse’s maximum instantaneous power output? Assume the same force of friction.

Express your answer using two significant figures.

P =

  hp  

Answer 1

part A

frictional force f = 850 N

If the sled moves up with constant velocity then the force applied by horse is same as frictional force

applied force F = 850 N

velocity v = 5 km/h = (5 * 1000) / 3600 m/s

v = 1.38 m/s

power P = F * v

P = 850 * 1.38

P = 1173 watt

We know

1 hp = 746 watt

Therefore power P = 1173 / 746 hp

P = 1.57 hp

part B

Given maximu velocity vmax = 18 km/h

vmax = 18 * 1000 / 3600 = 5 m/s

The acceleration produced is a = (v2 - v1) / t

a = (5 - 1.38) / 7

a = 0.517 m/s^2

The netforce is given by (applied force - frictional force)

F - f = m * a

F = f + (m * a)

F = 850 + (130 * 0.517)

F = 917.21 N

The maximum instantaneous power is

P = F * Vmax

P = 917.21 * 5

P = 4586.05 watt

P = 4586.05 / 746

P = 6.147 hp