Question

Constants

A 3.20 kg box is moving to the right with speed 8.00 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

What distance does the box move from its position at t=0 before its speed is reduced to zero?

Express your answer with the appropriate units.

If the force continues to be applied, what is the velocity of the box at 3.50 s ?

Express your answer with the appropriate units.

Answer 1

mass =m = 3.20 Kg Initial speed =u= 8.00 m/s to the right on surface frictionless

At t= 0 Horizontal force to left F = -6.00 N/s^2 x t^2 ( Force varying with time)

dv/dt = F/m = -6.00 t^2 /3.2 = -1.87 t^2

dv = -1.87 t^2 dt , If we integrate this from time 0 to time t

V-u = -1.87 t^3/3 = - 0.62 t^3, where V is velocity at any given time t

V = u - 0.62 t^3 --------- (1)

From (1) above , we get

dx = ( u-0.62 t^3) dt , on integrating this from time =0 to time =t

x = ut - 0.62 t^4/4 = 8.00 x t- 0.62x t^4/4 = 8t - 0.15 t^4

x = 8t- 0.15t^4 ---------- (2)

In (1) ,let us consider a time T at which velocity becomes = 0 m/s

V= 0 u = 8.00 m/s

-8.00 m/s = - 0.62T^3

T^3 = 8/0.62 = 12.9 sec^3

T = 2.34 secs

applying T = 2.34 in (2)

x = (8 x 2.34)- 0.15 ( 2.34)^4 = 14.2 m

To find velocity at t = 3.50 secs, we apply this in (1)

V = u - 0.62 t^3 = 8 m/s - (0.62 x 3.5^4 ) = - 85.0 m/s ( It is thus moving to left )