What impulse does the force shown in the figure exert on a 250 g particle?The graph is starts at (0, 1000). It is a constant line until (2, 1000) and then starts decreasing until the point...
What impulse does the force shown in the figure exert on a 250 g particle?
The graph is starts at (0, 1000). It is a constant line until (2, 1000) and then starts decreasing until the point (6,0). This is the point that touches the x-axis at (6,0) and the graph ends at this point. The y-axis is in terms of F_x (N) and the x-axis is in terms of miliseconds (ms).
1 Answer
valentin68 | College Teacher | (Level 3) Associate Educator
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`
