A 2.60 kg box is moving to the right with speed 10.0 m/s on a horizontal, fricti

Question

A 2.60 kg box is moving to the right with speed 10.0 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/s^2)t^2. If the force continues to be applied, what is the speed of the box at 3.00 s ?

Explanation / Answer

Force = -6t^2 ma = -6t^2 => a = -6t^2/2.6 dv/dt = -30t^2/13 integrate => v = -10t^3/13 + c at t =0 v= 10 => c =10 => v= -10t^3/13 + 10 for v = 0 -10t^3/13 + 10= 0 t = cube root(13) now v = dx/dt = -10t^3/13 + 10 => x(t) = -5t^4/26 +10t + c at t=0 x=0 => c = 0 => x = -5t^4/26 +10t now we need to find x at t= cube root(13) x = 17.63 distance the box moves is 17.63m

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