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Chapter 8: Problem 40

A girl on a skateboard (total mass of 40 kg) is moving at a speed of \(10 \mathrm{m} / \mathrm{s}\) at the bottom of a long ramp. The ramp is inclined at \(20^{\circ}\) with respect to the horizontal. If she uavels \(14.2 \mathrm{m}\) upward along the ramp before stopping. what is the net frictional force on her?

Short Answer

Expert verified
The net frictional force acting on the girl as she moves up the ramp is approximately \( -274.8\, \text{N} \). The negative sign indicates the force acts opposite to the direction of her motion (up the ramp).

Step by step solution

01

Calculate the initial kinetic energy of the girl

First, let's calculate the initial kinetic energy of the girl on the skateboard, using the formula for kinetic energy: \[K.E. = \frac{1}{2}mv^2\] where \(m\) is the mass of girl and skateboard combined (40 kg) and \(v\) is the initial speed (10 m/s). Plug in the values: \[K.E. = \frac{1}{2}(40 \,\text{kg})(10\, \text{m/s})^2 = 2000\, \text{J}\]
02

Calculate the total vertical height gained by the girl

Now, we will find the vertical height gained by the girl, using the following trigonometric relation: \[\text{Vertical height} = \text{distance} * \sin{\theta}\] where the distance traveled up the ramp is 14.2 m, and inclination angle \(\theta = 20^\circ\). Plug in the values: \[\text{Vertical height} = (14.2\, \text{m})\sin{20^\circ} \approx 4.85\, \text{m}\]
03

Calculate the gravitational potential energy gained by the girl

Now, let's calculate the gravitational potential energy gained by the girl, using the formula for potential energy: \[P.E. = mgh\] where \(m\) is the mass (40 kg), \(g\) is the acceleration due to gravity (9.81 m/s虏), and \(h\) is the vertical height (4.85 m). Plug in the values: \[P.E. = (40\, \text{kg})(9.81\, \text{m/s}^2)(4.85\, \text{m}) \approx 1902\, \text{J}\]
04

Apply the work-energy principle

According to the work-energy principle, the net work done on the girl is equal to the change in kinetic energy. Since she comes to a stop, her final kinetic energy is zero. So, the net work done, which involves the work done by friction, is equal to the negative of her initial kinetic energy: \[W_{net} = K.E._{final} - K.E._{initial} = -2000\, \text{J}\] However, the work done against gravity (gaining potential energy) is positive, so we can write: \[W_{net} = W_{friction} + 1902\, \text{J}\] and solve for the frictional work done: \[W_{friction} = -2000\, \text{J} - 1902\, \text{J} = -3902\, \text{J}\]
05

Calculate the net frictional force

Finally, we can calculate the net frictional force on the girl using the formula for work done: \[W_{friction} = F_{friction} \times d\] where \(F_{friction}\) is the frictional force, and \(d\) is the distance traveled up the ramp (14.2 m). Plug in the values and solve for the frictional force: \[F_{friction} = \frac{-3902\, \text{J}}{14.2\, \text{m}} \approx -274.8\, \text{N}\] The net frictional force acting on the girl as she moves up the ramp is approximately -274.8 N. The negative sign indicates the force acts opposite to the direction of her motion (up the ramp).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass of the object and its velocity. The formula for kinetic energy is:
  • \( K.E. = \frac{1}{2}mv^2 \)
  • Where \( m \) is the mass and \( v \) is the velocity.
In the original exercise, the girl on the skateboard, with a combined mass of 40 kg, was moving at 10 m/s. Her kinetic energy was calculated to be 2000 Joules (J). It is important to note that kinetic energy is directly proportional to the square of velocity, meaning that even small increases in speed can lead to significantly higher kinetic energy. The initial kinetic energy of the girl is a key factor in determining the work done by and against various forces, as observed in the exercise.
Potential Energy
Potential energy is the energy held by an object because of its position relative to other objects. In this scenario, as the girl moves up the ramp, she gains gravitational potential energy. The formula to calculate gravitational potential energy is:
  • \( P.E. = mgh \)
  • Where \( m \) is mass, \( g \) is acceleration due to gravity (approximately 9.81 m/s虏), and \( h \) is the height gained.
We found that the vertical height she gained was 4.85 meters. The potential energy corresponding to this height was approximately 1902 Joules. Potential energy is contingent on the height and gravitational forces involved. When the girl moved against gravity to a higher position on the ramp, her potential energy increased, impacting the overall energy balance in the system.
Trigonometry
Trigonometry plays a pivotal role in determining the vertical component of motion along an inclined plane. In physical calculations involving inclined planes, like ramps, trigonometric functions help resolve forces and distances into vertical and horizontal components. The formula used in these calculations is:
  • \( ext{Vertical height} = ext{distance} \times \sin(\theta) \)
  • Where \( \theta \) is the angle of inclination.
For the skateboard problem, a ramp inclined at 20掳 allowed us to compute the vertical height gained by the girl using the distance traveled up the ramp. This trigonometric approach to height calculation was essential for determining the gravitational potential energy.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. This principle is essential for analyzing scenarios involving force and motion. In the skateboard problem, it was crucial because it allowed us to calculate the net work done, which includes both the work done by friction and the work against gravitational forces. The principle can be expressed as:
  • \( W_{net} = K.E._{final} - K.E._{initial} \)
Applying this principle, the work against gravity was found to be positive and equal to the potential energy gained. The net work done was calculated by considering both this positive work and the negative work done by friction, balancing the forces that brought the girl to a stop. Through the work-energy principle, we arrived at a net frictional force of -274.8 N, indicating the opposition to her motion up the ramp.

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Most popular questions from this chapter

Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete's running kinetic energy to gravitational potential energy. If an athlete is to lift his body \(4.8 \mathrm{m}\) during a vault, what speed must he have when he plants his pole?

A child (32 kg) jumps up and down on a trampoline. The trampoline exerts a spring restoring force on the child with a constant of \(5000 \mathrm{N} / \mathrm{m}\). At the highest point of the bounce, the child is \(1.0 \mathrm{m}\) above the level surface of the trampoline. What is the compression distance of the trampoline? Neglect the bending of the legs or any transfer of energy of the child into the trampoline while jumping.

\(\begin{array}{lllll}\text { 52. } & \text { In } & \text { the } & \text { cartoon } & \text { movie } & \text { Pocahontas }\end{array}\) (https://openstaxcollege.org/l/21pocahontclip) Pocahontas runs to the edge of a cliff and jumps off, showcasing the fun side of her personality. (a) If she is running at \(3.0 \mathrm{m} / \mathrm{s}\) before jumping off the cliff and she hits the water at the bottom of the cliff at \(20.0 \mathrm{m} / \mathrm{s}\), how high is the cliff? Assume negligible air drag in this cartoon. (b) If she jumped off the same cliff from a standstill, how fast would she be falling right before she hit the water?

A mysterious constant force of 10 \(\mathrm{N}\) acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance \(x\) from the wall, assuming the potential energy at the wall to be zero.

A force \(F(x)=\left(-5.0 x^{2}+7.0 x\right) \mathrm{N}\) acts on a particle. (a) How much work does the force do on the particle as it moves from \(x=2.0 \mathrm{m}\) to \(x=5.0 \mathrm{m} ?\) (b) Picking a convenient reference point of the potential energy to be zero at \(x=\infty,\) find the potential energy for this force.
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