91Ó°ÊÓ

Cold Hot Dogs Figure \(9-32\) shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance \(d=\) \(20.0 \mathrm{~cm}\) while three forces are applied to it. Two of the forces are horizontal and have the magnitudes \(F_{A}=5.00 \mathrm{~N}\) and \(F_{B}=1.00 \mathrm{~N} ;\) the third force is angled down by \(\theta=-60.0^{\circ}\) and has the magnitude \(F_{C}=4.00 \mathrm{~N}\). (a) For the \(20.0 \mathrm{~cm}\) displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package? (b) If the package has a mass of \(2.0 \mathrm{~kg}\) and an initial kinetic energy of \(0 \mathrm{~J}\), what is its speed at the end of the displacement?

Step by step solution

01

Convert distance to meters

The distance given is in centimeters. Convert it to meters by dividing by 100.d = 20.0 cm / 100 = 0.20 m

02

Calculate the work done by each force

Use the formula for work: \( W = F \times d \times \text{cos}(\theta) \)- For the force \( F_{A} \):\( W_{A} = 5.00 \text{ N} \times 0.20 \text{ m} \times \text{cos}(0^\text{o}) = 1.00 \text{ J} \)- For the force \( F_{B} \):\( W_{B} = 1.00 \text{ N} \times 0.20 \text{ m} \times \text{cos}(180^\text{o}) = -0.20 \text{ J} \)- For the force \( F_{C} \):\( W_{C} = 4.00 \text{ N} \times 0.20 \text{ m} \times \text{cos}(-60^\text{o}) = 0.40 \text{ J} \)

03

Calculate the work done by gravitational and normal forces

Since the floor is frictionless and displacement is horizontal, both the vertical gravitational force and the normal force do no work as there is no vertical displacement.\( W_{\text{gravity}} = 0\text{ J} \)\( W_{\text{normal}} = 0\text{ J} \)

04

Calculate the net work done

Sum all the work values from the forces:\( W_{\text{net}} = W_{A} + W_{B} + W_{C} + W_{\text{gravity}} + W_{\text{normal}} \)\( W_{\text{net}} = 1.00 \text{ J} - 0.20 \text{ J} + 0.40 \text{ J} + 0 \text{ J} + 0 \text{ J} = 1.20 \text{ J} \)

05

Calculate the final speed

Use the work-energy principle which states that the net work done is equal to the change in kinetic energy. The initial kinetic energy is 0 J.\( W_{\text{net}} = \frac{1}{2} m v^2 - 0 \)Solving for the speed \( v \):\( 1.20 \text{ J} = \frac{1}{2} \times 2.0 \text{ kg} \times v^2 \)\( v^2 = 1.20 \text{ J} / (1 \text{ kg}) \)\( v = \text{sqrt}(1.20 \text{ m}^2/\text{s}^2) \)\( v = 1.10 \text{ m/s} \)

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Key Concepts

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

frictionless surface

Understanding how objects move on a frictionless surface is essential in physics. A frictionless surface is an idealized concept where there is no resistance to motion. In reality, almost every surface exhibits some degree of friction, but in many physics problems, assuming a frictionless surface simplifies the calculations and focuses on the primary forces at play.
When dealing with a frictionless surface:

• The only forces we consider are those explicitly mentioned in the problem, such as applied forces, gravitational force, and normal force.
• Since there's no friction, we don't have to worry about energy loss due to this force, making our calculations more straightforward.
• Objects on a frictionless surface continue moving in their current state (either at rest or in uniform motion) unless acted upon by an external force, as per Newton's First Law of Motion.

In our exercise, the cold package of hot dogs moves across a frictionless floor, meaning we only account for the applied forces and disregard any frictional force. This makes handling the calculation of work easier and helps us see the impact of each force clearly.

net work

The concept of net work is crucial in understanding how forces influence motion. Work is defined as the force applied on an object times the displacement of that object in the direction of the force. The formula for work is given by:

\[W = F \times d \times \text{cos}(\theta)\]

where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the displacement direction.
Net work is the total work done by all the forces acting on an object. In our exercise:

• The work done by each force is calculated by considering their magnitudes and angles.
• For forces along the direction of motion (\text{cos}(0\(^\text{o}\))), work is positive, and for forces opposite to the direction of motion (\text{cos}(180\(^\text{o}\))), work is negative.
• We sum up the work values to find the net work:
  
  \[W_{\text{net}} = 1.00 \text{ J} - 0.20 \text{ J} + 0.40 \text{ J} = 1.20 \text{ J}\]

This net work is what causes the change in the kinetic energy of the package, as per the work-energy principle.

kinetic energy

Kinetic energy is the energy an object possesses due to its motion. It is given by the formula:

\[KE = \frac{1}{2} mv^2\]

where \( m \) is the mass of the object, and \( v \) is its velocity.
Kinetic energy plays a key role in the work-energy principle, which states that the net work done on an object is equal to the change in its kinetic energy. In our exercise:

• The initial kinetic energy of the package is zero because it starts from rest.
• The net work done on the package (1.20 J) converts into kinetic energy.

Using the formula:

\[1.20 \text{ J} = \frac{1}{2} \times 2.0 \text{ kg} \times v^2\]
we find:

\[v^2 = \frac{1.20 \text{ J}}{1 \text{ kg}}\]
\[v = \text{sqrt}(1.20 \text{ m}^2/\text{s}^2)\]
\[v = 1.10 \text{ m/s}\]
Therefore, the speed of the package at the end of the displacement is 1.10 m/s. This demonstrates how work done by forces translates into the motion and energy of the object.