Help with work and vector forces

[Heisenburger]

Homework Statement

A particle of mass 1 kg moves under the force F = 5i + 15t^2j N, where t is in seconds. The initial velocity of the particle is 2i + 3j + 4k ms^−1. Find the work done by the force acting on the particle, and the increase in kinetic energy of the particle , during the time 0 ≤ t ≤ 1.

Homework Equations

F=ma or F=m(dv/dt)?

The Attempt at a Solution

i attempted to use the F=m(dv/dt) and put it equal to the force,

so 1(dv/dt)=5i+15t^2, then i integrated to get dx/dt=5ti+15/3*t^3,

I then added the initial velocity of 2i+3j+4k, so now had 5ti+15/3*t^3 +2i+3j+4k,

now i am thinking this is wrong, but my next plan was to integrate this again, with the limits of 0 to 1?

am i on the right track?

Thanks!

 

[Hollumber]

When you integrate 1(dv/dt), it should be from t = 0 to t=1. This gives you how much velocity is gained due to the force, and you can add this value to the initial velocity.

 

[Heisenburger]

When you integrate 1(dv/dt), it should be from t = 0 to t=1. This gives you how much velocity is gained due to the force, and you can add this value to the initial velocity.

So, would that tell me the final velocity?

could that then be used to work out kinetic energy?

what about the work?

Thanks!

 

[Hollumber]

Yes, adding the velocity due to the force to initial velocity gives final velocity.
Yes, this can be converted into KE.
Work is equal to the difference of KE's.

 

[Chestermiller]

Homework Statement

A particle of mass 1 kg moves under the force F = 5i + 15t^2j N, where t is in seconds. The initial velocity of the particle is 2i + 3j + 4k ms^−1. Find the work done by the force acting on the particle, and the increase in kinetic energy of the particle , during the time 0 ≤ t ≤ 1.

Homework Equations

F=ma or F=m(dv/dt)?

The Attempt at a Solution

i attempted to use the F=m(dv/dt) and put it equal to the force,

so 1(dv/dt)=5i+15t^2, then i integrated to get dx/dt=5ti+15/3*t^3,

I then added the initial velocity of 2i+3j+4k, so now had 5ti+15/3*t^3 +2i+3j+4k,

now i am thinking this is wrong, but my next plan was to integrate this again, with the limits of 0 to 1?

am i on the right track?

Thanks!

You are definitely on the right track. You can continue by first combining terms on the velocity so that you have the results in the form ai+bj+ck. The rate at which work is being done at any time is equal to the force F dotted with the velocity vector. Evaluate this dot product, and then integrate with respect to t. What is the formula for the change in KE in terms of the initial and the final velocity?

Chet

 

[Heisenburger]

Yes, adding the velocity due to the force to initial velocity gives final velocity.
Yes, this can be converted into KE.
Work is equal to the difference of KE's.

shall i leave the KE in vector form? or find the magnitude?

and for work do i do vector after-vector before then find magnitude or find the magnitudes and then take away?

THANKS!

 

[Chestermiller]

shall i leave the KE in vector form? or find the magnitude?

and for work do i do vector after-vector before then find magnitude or find the magnitudes and then take away?

THANKS!

Kinetic energy is a scalar, not a vector. KE is m(velocity dot velocity)/2. Calculate KE after and KE before, and then subtract.

 

[Heisenburger]

Kinetic energy is a scalar, not a vector. KE is m(velocity dot velocity)/2. Calculate KE after and KE before, and then subtract.

so, do i take the original equation for force, divide it by m (1) and integrate (with the limits of 1 & 0), to get the final velocity?

then FIRSTLY take the original velocity away from this new velocity, square that, multiply by m/2 to get KE.

Then SECONDLY take the original velocity, square it, multiply by m/2 then i have the original KE.

Then find out the difference, and that's the work?

 

[Chestermiller]

so, do i take the original equation for force, divide it by m (1) and integrate (with the limits of 1 & 0), to get the final velocity?

You already did that for arbitrary t.

then FIRSTLY take the original velocity away from this new velocity, square that, multiply by m/2 to get KE.

No. This is not the final KE. The final KE is just the final velocity squared multiplied by m/2. It does not depend on the initial velocity.

Then SECONDLY take the original velocity, square it, multiply by m/2 then i have the original KE.

Yes. So why didn't you suggest this same algorithm for the final KE?

Then find out the difference, and that's the work?

The difference is the work, but you're also supposed to show that the dot product of the force and the velocity integrated from t = 0 to t = 1 gives the identical result.

 

[Heisenburger]

You already did that for arbitrary t.

No. This is not the final KE. The final KE is just the final velocity squared multiplied by m/2. It does not depend on the initial velocity.Yes. So why didn't you suggest this same algorithm for the final KE?

The difference is the work, but you're also supposed to show that the dot product of the force and the velocity integrated from t = 0 to t = 1 gives the identical result.

Surely i have to take into account the initial velocity into my final KE value? as the force will have accelerated the particle but it still had the initial velocity to add on?

which is why i added the original velocity to the new derived one?

just realized earlier on, i meant ADD the initial velocity to the new equation, rather than TAKE IT AWAY which i stated?

is it now correct?

KE final = (New integrated velocity + initial velocity) * m/2?

Iniital is just initial * m/2?

 

[Chestermiller]

Surely i have to take into account the initial velocity into my final KE value? as the force will have accelerated the particle but it still had the initial velocity to add on?

No. Only the final velocity matters in the final kinetic energy.

which is why i added the original velocity to the new derived one?

just realized earlier on, i meant ADD the initial velocity to the new equation, rather than TAKE IT AWAY which i stated?

is it now correct?

KE final = (New integrated velocity + initial velocity) * m/2?

No. The final KE is the (final velocity)²m/2

 

[Hollumber]

Yes Heisen, your thinking is correct.

Final KE is (integrated V + initial V)^2 *m/2, since the integrated v only accounts for speed gained due to force and doesn't consider initial v (so we add it in afterwards).