Work and energy -- Change in KE due to a force F acting on a mass

In summary, the conversation is about a problem involving equations and a picture that is rotated by 90 degrees. The relevant equations are ##\vec F = m\vec a## and ##W = \int \vec F\cdot\vec s##. The person is trying to integrate ##\vec F = m\vec a## twice and is missing the given value of ##v_0##. They are also unable to find a solution to the issue with the rotated picture.
  • #1
Max2020
2
0
Homework Statement
a force of F = 3i + (6t ^ 2) j - tk acts on a particle of mass 2 kg, where F is given in newtons and t in seconds. if the initial velocity of the particle is Vo = j + 2k, in meters per second. () What is the work done by force F during the interval 0 <= t <= 2? () Using the definition of kinetic energy (K = mv ^ 2/2), find the kinetic energy variation ∆K of the particle in the same range.
Relevant Equations
F= 3i +(6t^2) j -tk
20210426_150244.jpg
 
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  • #2
Hello @Max2020 ,
:welcome: ##\qquad !##​

Did you notice your picture is rotated by 90 degrees ? It hurts to look at it !

Your $$\vec F= 3\,\hat\imath +6t^2 \,\hat\jmath -t\,\hat k $$ is not a relevant equation: it is part of the problem statement. Fortunately I do find a relevant equation (##\vec F = m\vec a##) in your picture if I almost break my neck.

The other relevant equation is ##W = \int \vec F\cdot\vec s##, also to be found in your work.

So you try to integrate ##\vec F = m\vec a\ ## twice. But I miss the given ##v_0 = \hat \jmath + 2\,\hat k\ ## there ?

Or do you have some other question that I somehow missed ?

##\ ##
 
  • #3
BvU said:
Hello @Max2020 ,
:welcome: ##\qquad !##​

Did you notice your picture is rotated by 90 degrees ? It hurts to look at it !

Your $$\vec F= 3\,\hat\imath +6t^2 \,\hat\jmath -t\,\hat k $$ is not a relevant equation: it is part of the problem statement. Fortunately I do find a relevant equation (##\vec F = m\vec a##) in your picture if I almost break my neck.

The other relevant equation is ##W = \int \vec F\cdot\vec s##, also to be found in your work.

So you try to integrate ##\vec F = m\vec a\ ## twice. But I miss the given ##v_0 = \hat \jmath + 2\,\hat k\ ## there ?

Or do you have some other question that I somehow missed ?

##\ ##
I'm sorry for posting the photo like this. I still can't see a way to resolve this issue.
 
  • #4
Max2020 said:
I'm sorry for posting the photo like this. I still can't see a way to resolve this issue.
1619481818607.png


:smile:
 

FAQ: Work and energy -- Change in KE due to a force F acting on a mass

What is the formula for calculating the change in kinetic energy due to a force?

The formula for calculating the change in kinetic energy (ΔKE) due to a force (F) acting on a mass (m) is ΔKE = ½mv², where v is the final velocity of the object.

How does the direction of the force affect the change in kinetic energy?

The direction of the force does not affect the change in kinetic energy. The change in kinetic energy depends only on the magnitude of the force and the final velocity of the object.

Can the change in kinetic energy be negative?

Yes, the change in kinetic energy can be negative. This means that the object's kinetic energy decreases due to the force acting on it. For example, if a force is applied in the opposite direction of the object's motion, the object's kinetic energy will decrease.

How does the mass of the object affect the change in kinetic energy?

The mass of the object does not directly affect the change in kinetic energy. However, a larger mass will require a greater force to produce the same change in kinetic energy as a smaller mass. This is because the change in kinetic energy is also dependent on the object's initial velocity.

What are some real-life examples of work and energy in action?

Some real-life examples of work and energy include a car accelerating, a person lifting a weight, a roller coaster going down a hill, and a ball rolling down a ramp. In all of these scenarios, a force is acting on an object, causing a change in its kinetic energy.

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