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Chapter 7: Problem 9

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
The work done on the canister is 20 J.

Step by step solution

01

Determine Initial and Final Velocities

The initial velocity of the canister is given as \(\vec{v}_i = 4.0 \ \mathrm{m/s}\) in the positive \(x\)-direction. The final velocity of the canister is \(\vec{v}_f = 6.0 \ \mathrm{m/s}\) in the positive \(y\)-direction.
02

Calculate Initial and Final Kinetic Energies

The initial kinetic energy (\(KE_i\)) of the canister is given by the formula \(\frac{1}{2}mv^2\). Plugging in the values, we have: \[ KE_i = \frac{1}{2} \times 2.0 \ \mathrm{kg} \times (4.0 \ \mathrm{m/s})^2 = 16 \ \mathrm{J} \]The final kinetic energy (\(KE_f\)) is calculated similarly: \[ KE_f = \frac{1}{2} \times 2.0 \ \mathrm{kg} \times (6.0 \ \mathrm{m/s})^2 = 36 \ \mathrm{J} \]
03

Determine the Work Done Using the Work-Energy Principle

The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy. We express this as:\[ W = KE_f - KE_i \]Substituting the kinetic energies from previous steps:\[ W = 36 \ \mathrm{J} - 16 \ \mathrm{J} = 20 \ \mathrm{J} \]

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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 possessed by an object due to its motion. It depends on two factors: the mass of the object and its velocity. The mathematical expression for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object (measured in kilograms).
  • \( v \) is the velocity of the object (measured in meters per second).
In the context of our exercise, the canister had an initial kinetic energy calculated with its initial velocity of 4.0 m/s. By inserting this into our formula along with the mass, we find the initial kinetic energy to be 16 Joules.
Later, the velocity changed to 6.0 m/s in a different direction, and so the final kinetic energy was calculated as 36 Joules with the same formula. The increase in kinetic energy indicates that work was done on the canister by an external force.
Newton's Second Law
Newton's Second Law of Motion plays a pivotal role in understanding how forces result in motion changes. It is formally stated as:\[ F = ma \]where:
  • \( F \) is the net force acting on an object (measured in newtons).
  • \( m \) is the mass of the object (measured in kilograms).
  • \( a \) is the acceleration produced in the object (measured in meters per second squared).
In our scenario, the canister was influenced by a force of 5.0 N, altering its velocity and thereby its kinetic energy. The change in velocity from the x-direction to the y-direction implies that the force applied to the canister caused it to accelerate and change its path. Applying Newton's Second Law helps us understand how the constant force is the reason behind the canister's change in speed and direction.
Vector Components
Vectors are quantities that have both magnitude and direction. In physics, many quantities like force or velocity are vectorial, including those in our exercise. When a vector is represented in a coordinate plane, it can be split into components along the x and y axes.
Initially, the canister's velocity was directed entirely in the x-direction, with a magnitude of 4.0 m/s. After some influence by the given force, the velocity changed direction completely to the y axis, reaching a magnitude of 6.0 m/s. This shift from one axis to another exemplifies how vector components can change under the influence of external forces.
Understanding vector components is crucial in decomposing any kind of motion or force within an x and y axis system, as it provides a clear picture of how objects move in a two-dimensional plane. It's essential also for calculating resultant vectors, which in cases like this, helps determine the path and changes in an object's motion.

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Most popular questions from this chapter

A particle travels through a three-dimensional displacement given by \(\vec{d}=(5.00 \hat{\mathrm{i}}-3.00 \hat{\mathrm{j}}+4.00 \hat{\mathrm{k}}) \mathrm{m}\). If a force of magnitude \(22.0 \mathrm{~N}\) and with fixed orientation does work on the particle, find the angle between the force and the displacement if the change in the particle's kinetic energy is (a) \(45.0 \mathrm{~J}\) and (b) \(-45.0 \mathrm{~J}\).

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 \(110 \mathrm{~N} / \mathrm{m}\). 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?

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(880 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(7.0\) \(\mathrm{m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0\) \(\mathrm{m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and (c) \(2.0 \mathrm{~m} / \mathrm{s}\) ?

A force \(\vec{F}=(3.00 \mathrm{~N}) \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}+(7.00 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a \(2.00 \mathrm{~kg}\) mobile object that moves from an initial position of \(\vec{d}_{i}=(3.00 \mathrm{~m}) \hat{\mathrm{i}}-(2.00 \mathrm{~m}) \hat{\mathrm{j}}+(5.00 \mathrm{~m}) \hat{\mathrm{k}}\) to a final position of \(\vec{d}_{f}=-(5.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}+(7.00 \mathrm{~m}) \hat{\mathrm{k}}\) in \(4.00 \mathrm{~s}\). Find (a) the work done on the object by the force in the \(4.00 \mathrm{~s}\) interval, (b) the average power due to the force during that interval, and (c) the angle between vectors \(\vec{d}_{i}\) and \(\vec{d}_{f}\).

Figure 7-34 shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an \(x\) axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height \(h=1.25 \mathrm{~m}\), so the cart slides from \(x_{1}=3.00\) \(\mathrm{m}\) to \(x_{2}=1.00 \mathrm{~m}\). During the move, the tension in the cord is a constant \(28.0 \mathrm{~N}\). What is the change in the kinetic energy of the cart during the move?
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