The graph of the force as a function of time is in the figure below.
There are two portions of the graph
`F_x(t) = 1000 N "for" t in[0,2] ms` ``
`F_x(t) =-250*t +1500 "for" t in [2,6] ms`
To compute the total impulse we start from the impulse theorem: Force is the variation of impulse `P` in the time unit.
`F =(dP)/dt` , or equivalent `dP = F*dt`
Now we integrate to find the total impulse
`P = int_0^PdP =int_0^0.006F(t)dt =int_0^0.002 1000*dt +int_0.002^0.006 (1500-250*t)*dt=`
`=1000*t (0->0.002) +[1500*t -(250/2)*t^2] (0.002 ->0.006) =`
`=1000*0.002 +1500*0.006-1500*0.002-125*0.006^2+125*0.002^2 =`
`=2+9-3-0.0045+0.0005 =7.996 kg*m/s`
Further Reading
We’ll help your grades soar
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support
Already a member? Log in here.
Are you a teacher? Sign up now