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(I) A \(16.0-\mathrm{kg}\) child descends a slide \(2.20 \mathrm{~m}\) high and reaches the bottom with a speed of \(1.25 \mathrm{~m} / \mathrm{s}\). How much thermal energy due to friction was generated in this process?

Short Answer

Expert verified
332.46 J of thermal energy was generated due to friction.

Step by step solution

01

Calculate Potential Energy at the Top

The potential energy (PE) can be calculated using the formula: \[ PE = mgh \]where:- \( m = 16.0 \, \text{kg} \) is the mass of the child,- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,- \( h = 2.20 \, \text{m} \) is the height of the slide.Plugging in the values:\[ PE = 16.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.20 \, \text{m} = 344.96 \, \text{J} \]
02

Calculate Kinetic Energy at the Bottom

The kinetic energy (KE) at the bottom is calculated using the formula:\[ KE = \frac{1}{2} mv^2 \]where:- \( m = 16.0 \, \text{kg} \) is the mass,- \( v = 1.25 \, \text{m/s} \) is the velocity at the bottom of the slide.Plugging in the values:\[ KE = \frac{1}{2} \times 16.0 \, \text{kg} \times (1.25 \, \text{m/s})^2 = 12.5 \, \text{J} \]
03

Determine Energy Lost to Friction

The energy lost to friction is the difference between the initial potential energy at the top and the kinetic energy at the bottom:\[ \text{Energy lost to friction} = PE - KE \]\[ \text{Energy lost to friction} = 344.96 \, \text{J} - 12.5 \, \text{J} = 332.46 \, \text{J} \]
04

Verify and Conclude

The thermal energy generated due to friction during the child's descent on the slide is found by calculating the difference between the initial potential energy and the kinetic energy at the bottom. This value is 332.46 Joules, meaning 332.46 J of thermal energy was generated.

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Key Concepts

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

Potential Energy
Potential energy is the energy an object possesses due to its position or state. It is stored energy that has the potential to do work. When an object is lifted against a gravitational field, it gains potential energy.
In our exercise, the child at the top of the slide represents an example of gravitational potential energy. This energy is calculated using the formula \[ PE = mgh \]Where:
  • \( m \) is the mass of the object (child in this case), which is \( 16.0 \text{ kg} \)
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \)
  • \( h \) is the height from which the object is elevated, \( 2.20 \text{ m} \)
The potential energy of the child at the top is calculated as \( 344.96 \text{ J} \). As the child descends the slide, this potential energy is transformed into other forms of energy, including kinetic and thermal energy.
Kinetic Energy
Kinetic energy is the energy of motion. When an object is in motion, it possesses kinetic energy, which depends on two key factors: mass and velocity. In our scenario, once the child reaches the bottom of the slide, they are moving, and hence they have kinetic energy.
The formula to calculate kinetic energy is: \[ KE = \frac{1}{2} mv^2 \]Where:
  • \( m \) is the mass of the object, which is \( 16.0 \text{ kg} \)
  • \( v \) is the velocity of the object at the bottom, \( 1.25 \text{ m/s} \)
Substituting these values gives the kinetic energy as \( 12.5 \text{ J} \).
This energy is much less than the initial potential energy because some energy has been transformed into other forms, primarily thermal energy due to friction.
Thermal Energy
Thermal energy refers to the energy that comes from heat. In mechanical processes, like a child sliding down a slide, energy is often partly converted to thermal energy, demonstrating the concept of energy dissipation.
In our exercise, as the child descends the slide, friction between the slide and the child generates heat, transforming some of the energy into thermal energy. This is considered energy loss since it is not available to do mechanical work.
The amount of thermal energy generated can be calculated by finding the difference between the potential energy at the top and the kinetic energy at the bottom:
  • \( 344.96 \text{ J} \) (initial potential energy) minus \( 12.5 \text{ J} \) (kinetic energy at the bottom) results in \( 332.46 \text{ J} \) of thermal energy generated due to friction.
Understanding these transformations and calculations highlights how energy changes forms yet is conserved in total, adhering to the principle of conservation of energy in physics.

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Most popular questions from this chapter

(1I) Draw a potential energy diagram, \(U\) vs. \(x,\) and analyze the motion of a mass \(m\) resting on a frictionless horizontal table and connected to a horizontal spring with stiffness constant \(k\) . The mass is pulled a distance to the right so that the spring is stretched a distance \(x_{0}\) initially, and then the mass is released from rest.

(II) A 1200 -kg car rolling on a horizontal surface has speed \(v=75 \mathrm{~km} / \mathrm{h}\) when it strikes a horizontal coiled spring and is brought to rest in a distance of \(2.2 \mathrm{~m}\). What is the spring stiffness constant of the spring?

(II) A 72 -kg trampoline artist jumps vertically upward from the top of a platform with a \(\begin{array}{llll}\text { speed of } & 4.5 \mathrm{~m} / \mathrm{s} . & (a) & \text { How }\end{array}\) fast is he going as he lands on the trampoline, \(2.0 \mathrm{~m}\) below (Fig. \(8-31\) )? (b) If the trampoline behaves like a spring of spring constant \(5.8 \times 10^{4} \mathrm{~N} / \mathrm{m}\), how far does he depress it?

(III) \(\mathrm{A} 2.0\) -kg block slides along a horizontal surface with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) The block has a speed \(v=1.3 \mathrm{m} / \mathrm{s}\) when it strikes a massless spring head- on. (a) If the spring has force constant \(k=120 \mathrm{N} / \mathrm{m},\) how far is the spring compressed? (b) What minimum value of the coefficient of static friction, \(\mu_{\mathrm{S}},\) will assure that the spring remains compressed at the maximum compressed position? (c) If \(\mu_{\mathrm{s}}\) is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detach- ment occurs when the spring reaches its natural length \((x=0) :\) explain why 1

(III) A \(2.0-\mathrm{kg}\) block slides along a horizontal surface with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) The block has a speed \(v=1.3 \mathrm{~m} / \mathrm{s}\) when it strikes a massless spring head- on (as in Fig. \(8-18\) ). \((a)\) If the spring has force constant \(k=120 \mathrm{~N} / \mathrm{m},\) how far is the spring compressed? (b) What minimum value of the coefficient of static friction, \(\mu_{S}\), will assure that the spring remains compressed at the maximum compressed position? (c) If \(\mu_{\mathrm{S}}\) is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detachment occurs when the spring reaches its natural length \((x=0) ;\) explain why.

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