AMU Medical AMU Solved Paper-2014

  • question_answer
    A block of mass 1.0 kg moving on a horizontal surface with speed 2m/s enters a rough surface. The retarding force \[({{F}_{r}})\] on the block is given by \[{{F}_{r}}=-\frac{k}{x};\,10\,m<x<100\,m\]\[=0\,;\,x<10\,m\] and \[x>100\,\,m\],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

    Correct Answer: D

    Solution :

                     The given, \[m=1.0\,kg,\,u=2\,m/s\]                 \[k=0.5\,J,\,x=100\,m\],                 v = ? Kinetic energy = ?                 \[{{F}_{r}}=-\frac{k}{x}=ma\] \[\Rightarrow \]               \[-\frac{0.5}{100}=1\times a\] \[\Rightarrow \]               \[a=-\frac{1}{200}m/{{s}^{2}}\]                 \[{{v}^{2}}={{u}^{2}}+2ax\]                 \[={{2}^{2}}-2\times \frac{1}{200}\times 100\]                 \[{{v}^{2}}=4-1\]                 \[{{v}^{2}}=3{{m}^{2}}/{{s}^{2}}\] Kinetic energy \[=\frac{1}{2}m{{v}^{2}}\]                                 \[=\frac{1}{2}\times 1\times 3=1.5\,J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner