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Chapter 9: Problem 43

Canister and One Force The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction, and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

Short Answer

Expert verified
20.0 J

Step by step solution

01

Identify the Given Data

The problem provides the following data: Mass of the canister: \( m = 2.0 \text{ kg} \), Initial velocity: \( \boldsymbol{v_i} = 4.0 \text{ m/s} \text{ in the positive } x \text{ direction} \), Final velocity: \( \boldsymbol{v_f} = 6.0 \text{ m/s} \text{ in the positive } y \text{ direction} \), Force magnitude: \( F = 5.0 \text{ N} \).
02

Calculate Initial and Final Kinetic Energy

Kinetic Energy (KE) is given by the formula \( KE = \frac{1}{2}mv^2 \). Calculate initial kinetic energy (KE_i): \[ KE_i = \frac{1}{2} \times 2.0 \text{ kg} \times (4.0 \text{ m/s})^2 = 16.0 \text{ J} \]. Calculate final kinetic energy (KE_f): \[ KE_f = \frac{1}{2} \times 2.0 \text{ kg} \times (6.0 \text{ m/s})^2 = 36.0 \text{ J} \].
03

Calculate the Work Done

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Calculate the change in kinetic energy (\( \Delta KE \)): \[ \Delta KE = KE_f - KE_i \]. Substitute the known values: \[ \Delta KE = 36.0 \text{ J} - 16.0 \text{ J} = 20.0 \text{ J} \]. Thus, the work done on the canister is \( 20.0 \text{ J} \).

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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. You can calculate it using the formula: \[ KE = \frac{1}{2}mv^2 \] where \(m\) is the mass of the object and \(v\) is its velocity.
Kinetic energy helps us understand how the energy of an object changes with its speed. In our exercise, the canister's kinetic energy changes as its velocity changes.
Initially, with a velocity of 4.0 m/s, its kinetic energy was calculated to be 16 J. Afterward, when the velocity increased to 6.0 m/s in a different direction, the kinetic energy increased to 36 J.
force magnitude
Force magnitude refers to the strength of a force acting on an object. It is measured in Newtons (N). In our problem, the canister is subjected to a force with a magnitude of 5.0 N.
The direction of this force causes the canister to change its motion over time. Understanding force magnitude is essential because it shows the capacity of the force to do work on the object.
In the exercise, this particular force is responsible for changing the kinetic energy of the canister by changing its velocity.
change in velocity
Change in velocity happens when an object鈥檚 speed or direction changes. It is a vector quantity, meaning it has both magnitude and direction.
Initially, the canister had a velocity of 4.0 m/s in the positive x-direction. Later, this velocity changed to 6.0 m/s in the positive y-direction.
This change in velocity is significant because it reflects the work done by the force on the canister.
To appreciate this, one must understand that velocity change directly impacts the kinetic energy, hence the work done.
work done
Work done is the energy transferred to or from an object by means of force causing a displacement. It鈥檚 calculated using the work-energy principle: \[ \text{Work Done} = \text{Change in Kinetic Energy} \ W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \]
In the exercise, we found the initial and final kinetic energies to be 16 J and 36 J, respectively.
The work done was then calculated to be 20 J, showcasing how energy input by force changes motion.
mass
Mass is the quantity of matter in an object and is measured in kilograms (kg). It plays a vital role in kinetic energy calculations and force interactions.
In our problem, the mass of the canister is constant at 2.0 kg. Understanding mass is important because it affects how much kinetic energy an object has for a given velocity.
A heavier object (greater mass) requires more energy to reach the same velocity as a lighter one.
velocity calculations
Velocity calculations are crucial to determining kinetic energy and understanding movement changes. Initial velocity (\(v_i\)) was given as 4.0 m/s, and final velocity (\(v_f\)) was 6.0 m/s.
We use these velocities to calculate kinetic energy: \[ KE_i = \frac{1}{2}mv_i^2 \] \[ KE_f = \frac{1}{2}mv_f^2 \]
By computing these, we can then find the change in kinetic energy and work done using the work-energy principle. Veering through initial and final velocities aids us in comprehending how force affects motion and energy.

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Most popular questions from this chapter

Spring at MIT During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m}\). If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

Floating Ice Block A floating ice block is pushed through a displacement \(\vec{d}=(15 \mathrm{~m}) \hat{\mathrm{i}}-(12 \mathrm{~m}) \hat{\mathrm{j}}\) along a straight embankment by rushing water, which exerts a force \(\vec{F}=(210 \mathrm{~N}) \mathrm{i}-(150 \mathrm{~N}) \hat{\mathrm{j}}\) on the block. How much work does the force do on the block during the displacement?

Velodrome (a) In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN}\), was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) \(\operatorname{In}\) 1960, Mrs. Maxwell Rogers of Tampa, Florida, reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm}\), how much work did her force do on the car?

Freight Car A railroad freight car of mass \(3.18 \times 10^{4} \mathrm{~kg}\) collides with a stationary caboose car. They couple together, and \(27.0 \%\) of the initial kinetic energy is transferred to nonconservative forms of energy (thermal, sound, vibrational, and so on). Find the mass of the caboose. Hint: Translational momentum is conserved.

Block at Rest A \(1.5 \mathrm{~kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an \(x\) axis is applied to the block. The force is given by \(\vec{F}(x)=\) \(\left(2.5 \mathrm{~N}-x^{2} \mathrm{~N} / \mathrm{m}^{2}\right) \mathrm{i}\), where \(x\) is in meters and the initial position of the block is \(x=0.0 \mathrm{~m} .(\) a) What is the kinetic energy of the block as it passes through \(x=2.0 \mathrm{~m} ?\) (b) What is the maximum kinetic energy of the block between \(x=0.0 \mathrm{~m}\) and \(x=2.0 \mathrm{~m}\) ?
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