A block of mass 1.0 kg moving on a horizontal surface with speed 2 m / s enters a rough surface. The retarding force (Fr) on the block is given by Fr= begincases-k / x ; 10 m

Question

A block of mass 1.0 kg moving on a horizontal surface with speed 2 m / s enters a rough surface. The retarding force (Fr) on the block is given by Fr= begincases-k / x ; 10 m< x< 100 m 0 ; x ; x<10 m text and x > 100 m endcases where, k=0.5 J. The kinetic energy of the block at x=100 m is (A) 4.5 J (B) 2.5 J (C) 0.5 J (D) 1.5 J

Tardigrade Admin · Accepted Answer

The given, m=1.0 kg , u=2 m / s k=0.5 J , x=100 m , v =? Kinetic energy =? Fr=-(k/x)=m a ⇒ -(0.5/100)=1 × a ⇒ a=-(1/200) m / s 2 v2=u2+2 a x =22-2 × (1/200) × 100 v2=4-1 v2=3 m2 / s 2 Kinetic energy =(1/2) m v2 =(1/2) × 1 × 3=1.5 J

Q. A block of mass 1.0kg moving on a horizontal surface with speed 2m/s enters a rough surface. The retarding force (Fr​) on the block is given by Fr​={−k/x;10m<x<100m0;x;x<10m and x>100m​ where, k=0.5J. The kinetic energy of the block at x=100m is

4.5 J

2.5 J

0.5 J

1.5 J

The given, m=1.0kg,u=2m/s k=0.5J,x=100m,v=? Kinetic energy =? Fr​=−xk​=ma ⇒−1000.5​=1×a ⇒a=−2001​m/s2 v2=u2+2ax =22−2×2001​×100 v2=4−1 v2=3m2/s2 Kinetic energy =21​mv2 =21​×1×3=1.5J

Q. A block of mass $1.0 k g$ moving on a horizontal surface with speed $2 m / s$ enters a rough surface. The retarding force $(F_{r})$ on the block is given by
$F_{r} = {- k / x; 10 m < x < 100 m 0; x; x < 10 m and x > 100 m$
where, $k = 0.5 J$ . The kinetic energy of the block at $x = 100 m$ is