Question

A horse draws a sled horizontally on snow at constant speed. The horse can produce a power of \(1.060 \mathrm{hp} .\) The coefficient of friction between the sled and the snow is \(0.115,\) and the mass of the sled, including the load, is \(204.7 \mathrm{~kg}\). What is the speed with which the sled moves across the snow?

Short answer

Question: Given a horse's power of 1.060 hp, a coefficient of friction between the sled and snow of 0.115, and a sled mass of 204.7 kg, find the speed at which the sled moves across the snow. Answer: The sled moves across the snow at approximately 3.43 m/s.

Step-by-step solution

Convert the given power to watts

Given the horse's power is 1.060 hp, we need to convert it to watts to solve the problem in SI units. We can use the following conversion factor: 1 hp ≈ 746 W Power in watts (P) = 1.060 hp × 746 W/hp ≈ 791.16 W

Calculate the force of friction acting on the sled

To calculate the force of friction (F_friction) acting on the sled, we can use the following formula: F_friction = μ × m × g where μ is the coefficient of friction, m is the mass of the sled, and g is the acceleration due to gravity (approximately 9.81 m/s^2). F_friction = 0.115 × 204.7 kg × 9.81 m/s^2 ≈ 230.89 N

Relate the power produced by the horse to the work done against friction

Now, we will relate the power produced by the horse to the work done against friction. The formula for power can be written as: P = F × v where P is the power produced by the horse, F is the force acting on the sled (which is equal to F_friction), and v is the speed of the sled.

Calculate the speed of the sled

We can rearrange the formula in step 3 to solve for the speed of the sled: v = P / F Substituting the values we calculated in steps 1 and 2: v = 791.16 W / 230.89 N ≈ 3.43 m/s Thus, the speed with which the sled moves across the snow is approximately 3.43 m/s.

Key concepts

Power Conversion

Understanding power conversion is essential in solving many physics problems, particularly when dealing with mechanical systems. In our exercise, we are given the power output of a horse in horsepower (hp) and need to convert it to watts (W) for use in calculations. Power measures how quickly work is done or energy is transferred. The conversion factor we use here, 1 hp approximately equal to 746 W, is a standard conversion in physics to shift from imperial units to the SI (International System of Units).

When solving problems, ensuring that units are consistent is crucial because equations are designed to work within a single unit system. The SI unit of power, the watt, is universally used in science and denotes a rate of energy usage per second. By converting to watts, we use a unit that aligns with other SI units, like newtons (N) for force and meters per second (m/s) for velocity, enabling us to solve our problem without unit discrepancies.

Force of Friction

Friction is the resisting force that occurs when two surfaces move or try to move across each other. It depends on the nature of the surfaces and the force pressing them together. In our case, friction plays a pivotal role as it is what the horse must overcome to pull the sled on snow.

The formula to calculate the force of friction is \( F_{\text{friction}} = \mu \times m \times g \), where \( \mu \) is the coefficient of friction, a unitless value representing the frictional properties between the two surfaces; \( m \) is the mass, and \( g \) is the acceleration due to gravity. This equation reflects the relationship between the frictional force and both the mass of the object being moved and the gravitational force acting upon it.

In our exercise, by calculating this force we understand the resistance the horse's power must overcome. Since the sled is moving at a constant speed, the force the horse applies due to its power equals the frictional force, indicative of a balance of forces as per Newton's first law of motion.

Work-Energy Principle

The work-energy principle is a pivotal concept in physics that links force, displacement, and energy. Essentially, it states that work done by forces will result in a change in kinetic energy. In the equation \( P = F \times v \), we relate the power (P) produced by the horse to the work done against the force of friction as the sled moves across the snow.

From this relationship, we understand that power is not just about the amount of energy transferred, but also the rate at which this energy is converted from one form to another. In the example problem, the steady speed of the sled indicates that the work done by the horse (via its power output) over time is being exactly balanced by the work done against friction - this is why the sled is moving at a constant velocity.

By rearranging the power equation, we can solve for speed (v), uncovering the rate at which the sled travels by the energy continuously outputted by the horse. It shows that if more power is exerted (assuming friction remains constant), the sled could move faster, illustrating the direct proportionality between power and velocity in this scenario.