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Frictional force is a force that resists the movement between two contacting surfaces. In the context of the sled problem, the frictional force opposes the direction of motion to keep the sled moving at a constant velocity. This frictional force is often referred to as kinetic friction because the sled is in motion. It depends on two main factors:

• The nature of the surfaces in contact
• The normal force which presses them together

For this exercise, the frictional force is equal to the horizontal component of the pulling force since the sled is moving at a constant velocity. This implies that there is no net acceleration.The equation relating frictional force (\( F_{f} \)) to the normal force (\( F_{n} \)) and the coefficient of kinetic friction (\( \mu \)) is: \[F_{f} = \mu \cdot F_{n}\] This shows that the frictional force is proportional to how hard the surfaces are pressed together, which is quantified by the normal force.

Identifying the horizontal component of a force involves breaking down the force acting at an angle into two perpendicular parts. In physics, especially in problems involving motion on horizontal surfaces, it's crucial to understand the influence of angled forces. For the 20.0-kg sled, the pulling force is directed at a 30-degree angle above the horizontal. To find how much of this force contributes to the sled's horizontal movement, we use the horizontal force component formula:\[F_{x} = F \cdot \cos(\theta)\]where \( F = 80.0 \mathrm{~N} \)and \( \theta = 30.0^{\circ} \).Substituting these values:\[F_{x} = 80.0 \cdot \cos(30.0^{\circ}) \approx 80.0 \cdot 0.866 \approx 69.3 \mathrm{~N}\]This calculation helps us understand the part of the pulling force that directly aids in moving the sled across the surface.

Normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface.In the sled problem, we need to account for the effect of the vertical component of the pulling force when calculating the normal force. Normally, you have:\[F_{g} = m \cdot g\]where \( m = 20.0 \mathrm{~kg} \)and \( g = 9.8 \mathrm{~m/s^2} \), giving\(F_{g} = 196.0 \mathrm{~N} \).However, since the pulling force is lifting the sled partly off the surface, \( F_{n} \) is reduced by the vertical component of the pulling force \( F_{y} \), calculated as:\[F_{y} = F \cdot \sin(30.0^{\circ}) = 40.0 \mathrm{~N}\]Therefore, the normal force acting on the sled is:\[F_{n} = F_{g} - F_{y} = 196.0 - 40.0 = 156.0 \mathrm{~N}\]This reduction is essential for correctly assessing the friction between the sled and the surface.

Trigonometry is indispensable in physics, as it allows us to resolve forces into their perpendicular components. This process aids in analyzing situations where forces act at angles rather than purely in one direction, offering a clearer view of the interactions involved.When dealing with a force at an angle, trigonometry helps in breaking down the force into horizontal and vertical components using sine and cosine functions. For an angle \( \theta \) with the horizontal:

• Cosine gives the horizontal component: \( F_{x} = F \cdot \cos(\theta) \)
• Sine gives the vertical component: \( F_{y} = F \cdot \sin(\theta) \)

For the sled example, this breakdown is critical in finding the effective forces acting parallel and perpendicular to the motion. These calculations are not only crucial for determining the net force affecting the sled but also for understanding how force at an angle distributes into influencing movement and opposing friction.