Force F-x=(10N)sin(2pit/4.0s) (where t in s ) is exerted on a 250 particle during the interval 0s<=t<=2.0s. If the particle starts from rest, what is its speed at t=2.0s?
I need to know how to do this as well as the answer. Thank you. :)
I need to know how to do this as well as the answer. Thank you. :)
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Is the mass 250g or 250kg or some other unit? Lets assume it's 250g = 0.25kg.
F = ma, so a = F/m = [(10)sin(2pi.t/4)] / 0.25
a = 40sin(2pi.t/4)
Since a = dv/dt, integrate with repsect to t to find v:
dv/dt = 40sin(2pi.t/4)
v = Integral [40sin(2pi.t/4)]dt
= -40(4/2pi)cos(2pi.t/4) + C
Particle starts from rest, so v=0 when t=0
0 = -40(4/2pi)cos(0) + C (note cos(0)=1)
C = 40(4/2pi) = 80/pi
When t=2s, v = -40(4/2pi)cos(2pi.2/4) + 80/pi
= -40(4/2pi)cos(pi)) + 80/pi (note cos(pi) = -1)
= 40(4/2pi) + 80/pi
= 160/pi = 51m/s
F = ma, so a = F/m = [(10)sin(2pi.t/4)] / 0.25
a = 40sin(2pi.t/4)
Since a = dv/dt, integrate with repsect to t to find v:
dv/dt = 40sin(2pi.t/4)
v = Integral [40sin(2pi.t/4)]dt
= -40(4/2pi)cos(2pi.t/4) + C
Particle starts from rest, so v=0 when t=0
0 = -40(4/2pi)cos(0) + C (note cos(0)=1)
C = 40(4/2pi) = 80/pi
When t=2s, v = -40(4/2pi)cos(2pi.2/4) + 80/pi
= -40(4/2pi)cos(pi)) + 80/pi (note cos(pi) = -1)
= 40(4/2pi) + 80/pi
= 160/pi = 51m/s