A 1200-kg car is being driven up a 5.0^∘ hill. The frictional force is directed opposite to the

step 1 So let's start by making a free body diagram. So we've got our car and there's the weight of the

A 1200-kg car is being driven up a 5.0^∘ hill. The...

A $1200-\mathrm{kg}$ car is being driven up a $5.0^{\circ}$ hill. The frictional force is directed opposite to the motion of the car and has a magnitude of $f=$ $524 \mathrm{N} .$ A force $\overrightarrow{\mathbf{F}}$ is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight $\overrightarrow{\mathbf{W}}$ and the normal force $\overrightarrow{\mathbf{F}}_{\mathrm{N}}$ directed perpendicular to the road surface. The length of the road up the hill is $290 \mathrm{m}$. What should be the magnitude of $\overrightarrow{\mathbf{F}},$ so that the net work done by all the forces acting on the car is $+150 \mathrm{kJ} ?$

A $1200-\mathrm{kg}$ car is being driven up a $5.0^{\circ}$ hill. The frictional force is directed opposite to the motion of the car and has a magnitude of $f=$ $524 \mathrm{N} .$ A force $\overrightarrow{\mathbf{F}}$ is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car:
its weight $\overrightarrow{\mathbf{W}}$ and the normal force $\overrightarrow{\mathbf{F}}_{\mathrm{N}}$ directed perpendicular to the road surface. The length of the road up the hill is $290 \mathrm{m}$. What should be the magnitude of $\overrightarrow{\mathbf{F}},$ so that the net work done by all the forces acting on the car is $+150 \mathrm{kJ} ?$

Key Concepts

Force Decomposition

When dealing with forces on an incline, it's important to decompose forces into components that are parallel and perpendicular to the surface. This technique simplifies the analysis by allowing one to consider the effect of each force component in the direction of motion, particularly the gravitational component that contributes to the work done along the incline.

Work-Energy Principle

This principle states that the net work done by all forces acting on a body is equal to the change in its kinetic energy. In problems involving motion, especially on an incline, it helps relate forces acting on the body, such as applied force, gravitational force, and friction, to the energy changes in the system.

Work Done by Forces

Work is defined as the component of a force in the direction of displacement multiplied by the displacement. This concept is crucial for understanding how different forces, when applied over a distance, contribute positively or negatively to the system's energy, ultimately affecting the motion and speed of an object.

Inclined Plane Dynamics

An inclined plane introduces a scenario where gravitational force can be decomposed into components parallel and perpendicular to the surface. This aspect of mechanics highlights the effects of gravitational pull along the incline, influencing the net work done on an object moving on a slope.

Friction

Friction is a resistive force that acts opposite to the direction of motion, resulting in a loss of mechanical energy. In dynamics problems, evaluating the work done by friction is vital since it subtracts from the net work, thereby affecting the net energy available for accelerating or maintaining the speed of an object.

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