A 1.50 kg box is moving to the right with speed 8.50 m/s on a horizontal, fricti

Question

A 1.50 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/S^2 ) t^2

1) What distance does the box move from its position at before its speed is reduced to zero?

2)If the force continues to be applied, what is the speed of the box at 3.00 ?

Explanation / Answer

1)acceleration = F/M =6/1.5=4m/sec^2 *t^2 acc=dv/dt => v=4*t^3/3 +c at time t= 0 v=-8.5m/sec =>c=-8.5 so v= 4*t^3/3-8.5.................1 v=dx/dt =>x=t^4/3 - 8.5*t +c1 at t=0, x=0 =>c1=0 so x=t^4/3 - 8.5*t................2 from 1 v= 0 at t=1.8542 sec x(1.8542)=11.82m (towards left from point where force starts acting) 2)v(t=3)=27.5m/sec towards left

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