Canister and One Force The only force acting on a 2.0 kg canister that is moving in an x y plan

Name: Problem 43, Chapter 9: Understanding Physics
Uploaded: 2021-08-04T20:19:50-08:00
Duration: 2 min 56 s
Channel: Shoukat Ali
Description: Canister and One Force The only force acting on a 2.0 kg canister that is moving in an x y plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m / s in the positive x direction, and some time later has a velocity of 6.0 m / s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time?

step 1 In this case, we have to calculate the work done by 5 .0 Newton force on the canister. That is W

Question

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?

   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?

Understanding Physics

Karen Cummings,… 1st Edition

Chapter 9, Problem 43

Instant Answer

Solved by Expert Shoukat Ali

08/04/2021

Step 1

Mathematically, this is represented as: \[ W = \Delta K \] where $ W $ is the work done, and $ \Delta K $ is the change in kinetic energy. Show more…

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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?

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Key Concepts

Work-Energy Theorem

This principle states that the net work done on an object is equal to the change in its kinetic energy. It provides a direct relation between the forces applied to an object and the resulting change in the object's speed, regardless of the path taken.

Kinetic Energy

Kinetic energy is the energy that an object has due to its motion, defined by the equation (1/2)mv². Changes in an object's speed alter its kinetic energy, and these changes can be directly computed to assess work done.

Force as a Vector

Forces are vector quantities, meaning they have both magnitude and direction. When analyzing motion, especially in problems involving changes in direction, it is crucial to consider the vector nature of force to correctly apply principles such as the Work-Energy Theorem.

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