Question

5.23 CP CALC A 2.10-kg box is moving to the right with speed 8.50 m/s on a horizontal, frictionless surface. At t 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t) (6.00 N/s2)12. (a) What distance does the box move from its position at t0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at 3.00 s?

Answer 1

answer) a) first of all we will determine the time when the object comes to rest

taking integration

$\int_{0}^{t}$-Fdt=mvf-mvi=$\int_{0}^{t}$-6t²dt=0-mvo

2t³=2.1*8.5

t³=17.85/2=8.925

t=2.074s

now using work and energy relation we have

$\Delta$k=F*d

Fd=1/2m(vf-vi)²

d=-mvi²/F

d=2.1*8.5²/$\int_{0}^{2.074}$-6t²

d=151.725/2*2.074³=8.50 m

so answer is 8.50 m

b) now again integrating

$\int_{0}^{3}$-6t²dt=mvf-mvi

2t³|₀³=2.1vf-2.1*8.5

-54=2.1vf-17.85

vf=-36.15/2=-18.075m/s

so the answer is -18.1 m/s