Question

9) A force in the +x-direction with magnitude F(x) = 18.0 N – (0.530 N/m) x is applied to a 6.00-kg box that is sitting on a frictionless surface. F(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?

Answer 1

Fx = 18-0.53N/m * 14m = 10.58 N.

a = F/m = 10.58/6 = 1.76 m/s^2.

V^2 = Vo^2 + 2a*d = 0 + 2*1.76*14 = 49.4
V = 7.03 m/s.

Answer 2

this is wrong

Answer 3

Take the integral of F(x) which is

18.0x-.5(.530)x^2 then plug in 14.0 for x
=200.06J
so .5mv^2=W since v(initial) is zero so then solve for v
v(final)=((2*200.06)/6)^(1/2)
v(final)=8.17m/s

Answer 4

Sir mirian. May I ask why is their a .5 next to (0.53)?

Answer 5

when you integrate x you get (x^2)/2, the .5 represents the 1/2