CP CALC? A 2.00-kg box is moving to the right with speed 9.00 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)t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
Solution 23E According to newton’s second law F=ma The force is not constant, so the acceleration is not constant,so we cannot use the kinematic equations. we have to use newton’s second law The box is moving right side so its direction is toward positive x axis F x ax(t)= m (6 N/s )t = 2 kg =(-3 m/s )t 2 To find the velocity as a function of time, integrate the acceleration 4 4 vx(t)=-(1m/s )t +9 m/s To find the velocity as a position of time, integrate the velocity 4 3 x(t)=-(1 m/s )(3 s) +(9 m/s)t a) vx=0 when (1m/s )t =9 m/s 3 9 m/s t = (1m/s ) t= 2.08 s At t= 2.08 s 4 x(t)=(9 m/s)(2.08 s) -(0.250 m/s )=18.72 m-4.68 m=14 m b) At t=3 s 4 3 vx(t)=-(1m/s )(3 s) +9 m/s =-18 m/s The box starts out moving to the right. But because the acceleration is to the left.
