I really need some help with this one, I can\'t seem to figure it out. An object

Question

I really need some help with this one, I can't seem to figure it out.

An object with a mass of 4 kg travels with a constant velocity of 5.8 m/s northward. It is then acted on by a 6 Newton force in the direction of motion and a force of 9.5 Newton to the South, both of which continue even after the mass comes momentarily to rest.

How far will the object travel before coming to rest?

What will be its position 1.5seconds after the object comes momentarily to rest?

Then would this work correctly?

a = F/m = 3.5/4 = .875 m/s^2
t = 1.66 sec
V = 5.8 m/s

Distance = Vt + 1/2at^2
Distance = (5.8)(1.66) + 1/2(-.875)(1.66)^2
Distance = 8.42 meters

I feel like I am setting this up wrong. Any help would be greatly appreciated.

Explanation / Answer

mass of the object = 4kg velocity of body vi = 5.8m/s final velocity vf = 0m/s resultant force = 9.5N-6.0N =3.5N acceleration = 3.5/4=0.875m/s^2 it is deceleration = -0.875m/s^2 distance travelled before stop s=? (vf)2=(vi)2+2(-a)s s=(vi)2/2a s=(5.8*5.8)/2*0.875 s=33.64/1.75 s=19.22m the distance traveled after 1.5sec = s1 s1=(vi)t+(1/2)(-a)t2 s1=(5.8*1.5)+(1/2)(-0.875)(1.5*1.5) s1=8.7-0.98 s1=7.72m s1=(5.8*1.5)+(1/2)(-0.875)(1.5*1.5) s1=8.7-0.98 s1=7.72m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---