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Hi Umang, The correct ans is 2ax[1+(x/r)^2]^1/2 mv2/2=ax2 So v=(sqrt(2a/m))x Tangential acceleration a1=dv/dt==(sqrt(2a/m))dx/dt==(sqrt(2a/m))v=2ax/m Centripetal acc a2=v2/r=2ax2/mr resultant acc A=sqrt(a1square+a2square) Force=mA =2ax[1+(x/r)^2]^1/2 This is the correct ans and I am 300% sure. |
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-ADARSH NITK Surathkal |
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