How to Calculate Velocity and Position from a Force-Time Graph?

[gasar8]

We are pulling a box with a mass = 8kg up a slope (ψ=30°). Force of a hand is given F(t)=F₀*sin(Ωt). F₀= 150N, Ω=(2*∏)/t₀, t₀=30s. After t_f=15s we don't pull it anymore.
a) Find average velocity at [t=12s to t=10s], current velocity at t=11s and v(t) before and after the end of pulling.
b) Find x(t).

I am having problems to find velocity ot of force function. Can anyone help me please? :)
Thank you in advance. :)

 

[NihalSh]

show your work.

 

[gasar8]

Ok. First I calculated Fd (the component of Fg down the slope => sin30°*Fg=40N). Then I calculated the F(t=10s)=129,9N and F(t=12s)=88,17N. Then I figured out that the area under the graph F(t) = integral = F*Δt=m*Δv. So F*Δt= 221,32Ns. Then I divided by 2s (Δt) so that I got Force alone on one side. I got 110,66N. Now, I substracted Fd from this F and got 70,66N=(m*Δv)/Δt. From that equation I got Δv=17,67 m/s, but from now on I couldn't find velocities out of the Δv.
I feel terrible, because I stared in this exercise for about 4 hours and I tried everything, always got different results.

 

[Tanya Sharma]

Hi gasar8

Welcome to Physicsforums!

We are pulling a box with a mass = 8kg up a slope (ψ=30°). Force of a hand is given F(t)=F₀*sin(Ωt). F₀= 150N, Ω=(2*∏)/t₀, t₀=30s. After t_f=15s we don't pull it anymore.
a) Find average velocity at [t=12s to t=10s], current velocity at t=11s and v(t) before and after the end of pulling.
b) Find x(t).

I am having problems to find velocity ot of force function. Can anyone help me please? :)
Thank you in advance. :)

The force by hand is a sinusoidally varying force given by F=150Sin(2πt/30).

The net force along the incline is F - mgsin30°

So, 150Sin(2πt/30) - 40 = 8dv/dt

Integrate with proper limits and you will get velocity as a function of time v=f(t) .From this you can get velocity at 11s and velocity before pulling.Similarly you can get an expression for velocity after pulling stops.

Again write v= dx/dt = f(t) .Integrate with proper limits and you have displacement as a function of time . Say x(t) =g(t)

Now v_avg = [x(t₂)-x(t₁)]/(t₂-t₁) =[x(12)-x(10)]/2

Just give it a try.

 

[gasar8]

Thank you, but I still don't know if I think in a right way.
So, I wrote dv= (150sin((2∏t)/30)-40)/8 dt
∫dv (from v₀ - v)=∫((from 0-t) (150sin((2∏t)/30)-40)/8) dt
If a is not constant do I have to integrate it, too?
If yes, I get:
v= ((-2250*cos((∏t)/15))/∏ -40t)/8
If I place t=11s I get v=4,9 m/s

Fot v_avg I calculated v(t=10s) and v(t=12s) and got 3,6m/s. What is strange is, that v(t=10s)= - 5,24m/s.

I really don't understand this concept well. -.-

 

[Tanya Sharma]

Thank you, but I still don't know if I think in a right way.
So, I wrote dv= (150sin((2∏t)/30)-40)/8 dt
∫dv (from v₀ - v)=∫((from 0-t) (150sin((2∏t)/30)-40)/8) dt
If a is not constant do I have to integrate it, too?
If yes, I get:
v= ((-2250*cos((∏t)/15))/∏ -40t)/8
If I place t=11s I get v=4,9 m/s

Fot v_avg I calculated v(t=10s) and v(t=12s) and got 3,6m/s. What is strange is, that v(t=10s)= - 5,24m/s.

I really don't understand this concept well. -.-

150Sin(2πt/30) - 40 = 8dv/dt

[150Sin(2πt/30) - 40]dt = 8dv

dv = (150/8)Sin(2πt/30)dt - (40/8)dt

Integrate on both the sides .What do you get ?

Note:Assuming the box starts from rest,the initial velocity is zero.

 

[gasar8]

Hi,

I said that ∫(from 0 to 11) dv = ∫(from 0 to 11) (150/8)Sin(2πt/30)dt - (40/8)dt and I got (1125(1+sin7∏/30))/4∏ - 55 from that v(t=11s)=94,4m/s which is very strange to me. -.-
I really can't find the solution. Can you please explain this exercise to me to the end, I must hand it in tomorrow.

 

[gasar8]

Hello, today I used wolfram alpha to integrate and I got:
v(t)= (-F_dtω-F₀cos(tω)+F₀)/(ω*m)...v(11s)=94,43m/s
After pulling: -F_d*t/m

x(t)= (tω(2F₀-F_dtω)-2F₀sin(tω))/(2mω²)
x(11s)=364,6m
After pulling: -F_d*t²/(2m)
x(40s)=-4km

v_avg=Δx/Δt...463-275/2=94m/s

I asked my friend for help and he got the same, but those results are very strange. ?!

 

[Tanya Sharma]

Hi,

I said that ∫(from 0 to 11) dv = ∫(from 0 to 11) (150/8)Sin(2πt/30)dt - (40/8)dt and I got (1125(1+sin7∏/30))/4∏ - 55 from that v(t=11s)=94,4m/s which is very strange to me. -.-
I really can't find the solution. Can you please explain this exercise to me to the end, I must hand it in tomorrow.

Hello, today I used wolfram alpha to integrate and I got:
v(t)= (-F_dtω-F₀cos(tω)+F₀)/(ω*m)...v(11s)=94,43m/s
After pulling: -F_d*t/m

x(t)= (tω(2F₀-F_dtω)-2F₀sin(tω))/(2mω²)
x(11s)=364,6m
After pulling: -F_d*t²/(2m)
x(40s)=-4km

v_avg=Δx/Δt...463-275/2=94m/s

I asked my friend for help and he got the same, but those results are very strange. ?!

I simply can't follow you .You need to have more clarity while expressing your work . I am assuming you have written v(t=11s)=94.4m/s and not 94,4m/s . Why does it look strange to you ? The answer is what you get mathematically as per the question .

What prompts you to calculate x(40s) ?

Anyways, we have

dv = (150/8)Sin(2∏t/30)dt - (40/8)dt

∫dv = ∫(150/8)Sin(2∏t/30)dt - ∫5dt

v(t) = -(4500/16∏)Cos(2∏t/30) - 5t + C ,where C is constant of integration

Now ,v(0) = 0 ,gives C = 4500/16∏

So,v(t) = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

Now write v(t)=dx/dt ,

dx/dt = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

dx = -(4500/16∏)Cos(2∏t/30) - 5t + 4500/16∏

∫dx = -∫(4500/16∏)Cos(2∏t/30)dt - ∫(5t)dt + ∫(4500/16∏)dt

What value of x(t) do you get ? Please do not plug in the numbers .Just write down your expression of x(t) clearly .

 

[NihalSh]

Hello, today I used wolfram alpha to integrate and I got:
v(t)= (-F_dtω-F₀cos(tω)+F₀)/(ω*m)...v(11s)=94,43m/s
After pulling: -F_d*t/m

x(t)= (tω(2F₀-F_dtω)-2F₀sin(tω))/(2mω²)
x(11s)=364,6m
After pulling: -F_d*t²/(2m)
x(40s)=-4km

v_avg=Δx/Δt...463-275/2=94m/s

I asked my friend for help and he got the same, but those results are very strange. ?!

Like Tanya already mentioned, if that is the result that comes out after correctly solving it then why is it strange?...if you have specific doubt regarding the result, post it. BTW avg velocity does come out to be 94m/s.