Work On Inclined Planes: Videos & Practice Problems

To calculate the work done by gravity on an object on an inclined plane, break down the gravitational force (mg) into components: mgx (parallel to the incline) and mgy (perpendicular). Use the equation for work, W = Fd × cosθ, where the angle is between the force and displacement. For mgx, the work is 7,080 joules, while mgy does no work (0 joules) since it acts perpendicular to the motion. Thus, the total work done by gravity equals the work done by mgx.

Work By Gravity On Inclined Planes

In problems involving an object on an inclined plane, calculating the work done by gravity can be challenging due to the need to break down gravitational force into its components. Consider a scenario where a 100 kg block slides down an incline at a 37-degree angle over a distance of 12 meters. To find the work done by the gravitational force, we need to analyze the components of the gravitational force, denoted as \( mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

The gravitational force can be decomposed into two components: \( mg_x \) (the component parallel to the incline) and \( mg_y \) (the component perpendicular to the incline). The work done by a force is calculated using the formula:

\( W = F \cdot d \cdot \cos(\theta) \)

Here, \( W \) is the work done, \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.

For the component \( mg_x \), which acts down the ramp, the angle \( \theta \) between \( mg_x \) and the displacement is 0 degrees. Thus, the work done by \( mg_x \) can be calculated as:

\( W_{mg_x} = mg_x \cdot d \cdot \cos(0) = mg_x \cdot d \)

Substituting the values, we find:

\( mg_x = mg \cdot \sin(\theta_x) = 100 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot \sin(37^\circ) \)

Calculating this gives:

\( W_{mg_x} = 100 \cdot 9.8 \cdot \sin(37^\circ) \cdot 12 \cdot 1 = 7,080 \, \text{J} \)

On the other hand, the component \( mg_y \) acts perpendicular to the displacement. The angle between \( mg_y \) and the displacement is 90 degrees, leading to:

\( W_{mg_y} = mg_y \cdot d \cdot \cos(90) = 0 \, \text{J} \)

This indicates that the work done by the perpendicular component of gravity is always zero, as it does not contribute to the motion along the incline.

To find the total work done by gravity, we can sum the work done by both components:

\( W_{mg} = W_{mg_x} + W_{mg_y} = 7,080 \, \text{J} + 0 \, \text{J} = 7,080 \, \text{J} \)

Thus, the work done by gravity on the block sliding down the incline is \( 7,080 \, \text{J} \). This illustrates that the work done by gravity in inclined plane scenarios is primarily due to the component of gravitational force acting parallel to the incline, reinforcing the importance of correctly identifying the angles and components involved in the calculations.

Pulling Crate Up An Inclined Plane

In this scenario, we analyze the motion of a crate being pulled up an inclined plane. The mass of the crate is given as 19 kilograms, and an applied force of 130 newtons acts parallel to the ramp. The gravitational force acting on the crate can be expressed as \( mg \), where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \). To simplify calculations, we resolve the gravitational force into two components: one acting down the incline (\( mg_x \)) and the other perpendicular to the incline.

To calculate the work done by gravity as the crate moves up the incline, we use the formula for work done by a force, which is given by \( W = F \cdot d \cdot \cos(\theta) \). In this case, the work done by gravity can be expressed as:

\[ W_{mg} = -mg \sin(\theta) \cdot d \]

Here, \( \theta \) is the angle of the incline, which is 36 degrees, and \( d \) is the distance moved along the incline, given as 15 meters. Plugging in the values, we find:

\[ W_{mg} = - (19 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot \sin(36^\circ) \cdot 15 \, \text{m} \approx -1640 \, \text{joules} \]

The negative sign indicates that gravity is doing negative work as the crate is pulled up the ramp.

Next, we apply the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. The net work is the sum of the work done by all forces acting on the crate. In this case, we consider the work done by the applied force and the work done by gravity. The work done by the applied force can be calculated as:

\[ W_{FA} = F_A \cdot d \cdot \cos(0^\circ) = 130 \, \text{N} \cdot 15 \, \text{m} = 1950 \, \text{joules} \]

Since the applied force and the displacement are in the same direction, the angle is 0 degrees, and thus \( \cos(0^\circ) = 1 \). Now, we can find the net work:

\[ W_{net} = W_{FA} + W_{mg} = 1950 \, \text{joules} - 1640 \, \text{joules} = 310 \, \text{joules} \]

Since the crate starts from rest, the initial kinetic energy is zero. Therefore, the final kinetic energy of the crate is equal to the net work done:

\[ KE_{final} = W_{net} = 310 \, \text{joules} \]

This analysis illustrates the relationship between work, energy, and motion on an inclined plane, highlighting how forces interact to influence the kinetic energy of an object.

Finding Speed of a Crate Using Work-Energy Theorem

In this scenario, we analyze a crate sliding down an inclined plane, focusing on the forces acting on it and calculating the work done by gravity and friction, as well as the final speed of the crate. The crate has a mass of 7 kg and slides down a ramp at an angle of 26 degrees over a distance of 2.5 meters.

First, we identify the forces acting on the crate: the gravitational force (\(mg\)), the normal force, the component of gravitational force acting down the ramp (\(mg_x\)), and the frictional force opposing the motion. The work done by gravity (\(W_{mg}\)) can be calculated using the formula:

\(W_{mg} = mg \sin(\theta) \cdot d\)

Here, \(m\) is the mass (7 kg), \(g\) is the acceleration due to gravity (approximately 9.8 m/s²), \(\theta\) is the angle of the incline (26 degrees), and \(d\) is the distance slid (2.5 m). Plugging in the values, we find:

\(W_{mg} = 7 \cdot 9.8 \cdot \sin(26^\circ) \cdot 2.5 \approx 75.2 \, \text{J}\)

This positive value indicates that gravity is doing work on the crate, increasing its speed as it slides down the ramp.

Next, we calculate the work done by friction (\(W_f\)). The work done by friction is always negative, as it opposes the motion. The formula for work done by friction is:

\(W_f = -f_k \cdot d\)

Where \(f_k\) is the kinetic friction force, which can be expressed as:

\(f_k = \mu_k \cdot N\)

On an inclined plane, the normal force (\(N\)) is given by:

\(N = mg \cos(\theta)\)

Thus, the work done by friction becomes:

\(W_f = -\mu_k \cdot (mg \cos(\theta)) \cdot d\)

Assuming a coefficient of kinetic friction (\(\mu_k\)) of 0.36, we can calculate:

\(W_f = -0.36 \cdot (7 \cdot 9.8 \cdot \cos(26^\circ)) \cdot 2.5 \approx -55.5 \, \text{J}\)

This negative value reflects the energy lost due to friction as the crate slides down the ramp.

To find the final speed of the crate, we apply the work-energy theorem, which states that the net work done on the crate equals the change in kinetic energy:

\(W_{net} = \Delta KE = KE_{final} - KE_{initial}\)

Since the crate starts from rest, \(KE_{initial} = 0\), so:

\(W_{net} = KE_{final}\

Calculating the net work:

\(W_{net} = W_{mg} + W_f = 75.2 - 55.5 = 19.7 \, \text{J}\)

Setting this equal to the final kinetic energy:

\(19.7 = \frac{1}{2} mv_{final}^2\

Substituting the mass:

\(19.7 = \frac{1}{2} \cdot 7 \cdot v_{final}^2\

Solving for \(v_{final}\), we rearrange the equation:

\(v_{final}^2 = \frac{19.7 \cdot 2}{7}\

Calculating gives:

\(v_{final}^2 \approx 5.6286\

Taking the square root yields:

\(v_{final} \approx 2.4 \, \text{m/s}\

This final speed indicates how fast the crate is moving at the bottom of the incline after accounting for the work done by gravity and the work done against friction.

Here’s what students ask on this topic:

How do you calculate the work done by gravity on an inclined plane?

To calculate the work done by gravity on an inclined plane, you need to break down the gravitational force (mg) into two components: mg_x (parallel to the incline) and mg_y (perpendicular to the incline). The work done by gravity is primarily due to the mg_x component. Use the equation for work, $W=Fd\times cos\theta$, where θ is the angle between the force and displacement. For mg_x, the work is calculated as $W=\mathrm{mg}\times sin\theta \times d\times cos0$. Since cos(0) = 1, the work done by mg_x is $W=\mathrm{mg}\times sin\theta \times d$. The mg_y component does no work as it is perpendicular to the motion.

What is the role of the angle in calculating work on an inclined plane?

The angle plays a crucial role in calculating work on an inclined plane. In the equation $W=Fd\times cos\theta$, θ is the angle between the force and the displacement. For the component of gravity parallel to the incline (mg_x), θ is 0 degrees because the force and displacement are in the same direction. For the perpendicular component (mg_y), θ is 90 degrees, making the work done by mg_y zero since cos(90) = 0. It's important not to confuse the incline angle with the angle between force and displacement.

Why does the perpendicular component of gravity (mg_y) do no work on an inclined plane?

The perpendicular component of gravity (mg_y) does no work on an inclined plane because it acts perpendicular to the direction of motion. Work is defined as $W=Fd\times cos\theta$, where θ is the angle between the force and displacement. For mg_y, this angle is 90 degrees, and cos(90) = 0. Therefore, the work done by mg_y is zero. This makes sense because mg_y does not contribute to the motion along the incline.

How do you determine the angle between force and displacement on an inclined plane?

To determine the angle between force and displacement on an inclined plane, you need to analyze the direction of each component of the gravitational force relative to the motion. For the parallel component (mg_x), the force and displacement are in the same direction, so the angle θ is 0 degrees. For the perpendicular component (mg_y), the force is perpendicular to the displacement, making the angle θ 90 degrees. Always ensure you are using the angle between the force vector and the direction of motion, not the incline angle itself.

What is the total work done by gravity on an inclined plane?

The total work done by gravity on an inclined plane is the sum of the work done by its components, mg_x and mg_y. Since mg_y does no work (0 joules) because it is perpendicular to the motion, the total work done by gravity is equal to the work done by mg_x. This can be calculated using $W=\mathrm{mg}\times sin\theta \times d$. For example, if a 100 kg block slides down a 37-degree incline for 12 meters, the work done by gravity is 7,080 joules.