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A man moves on a straight horizontal road with a block in his hand . If he covers a distance of 40m with an acceleration of 0.5ms^-2. find the work done during the motion
assume the mass of block be m so work done can be calculated W =change in kinetic energy =K2-K1the velocity after covering 40m V2=u2+2aSSo V2 =40Now kinetic energy will be (1/2)mV2 =20mSo the work done will be 20mJ
Let suppose a block of mass m so work energy theoryWork done=change in kinetic energy =K2-K1the velocity attined after covering 40m V2=u2+2aSSo V2 =40Now change in kinetic energy will be (1/2)mV2 =20mSo the work done will be 20mJ using above equations
Given that a = 0.5 ms^2m = 2 kgs = 40 m Now to find v^2 we usev^2= u^2 + 2asv= √40Since by work energy theromWork done=∆ K.E.W= 40J
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