A horizontal cable pulls a 200-kg cart along a horizontal track. The tension in the cable is 500 N. Starting from rest, (a) How long will it take the cart to reach a speed of 8.0 m]s? (b) How far will it have gone? A horizontal cable pulls a 200-kg cart along a horizontal track. The tension in the cable is 500 N. Starting from rest, (a) How long will it take the cart to reach a speed of 8.0 m]s? (b) How far will it have gone?

Question

A horizontal cable pulls a 200-kg cart along a horizontal track. The tension in the cable is 500 N. Starting from rest, (a) How long will it take the cart to reach a speed of 8.0 m]s? (b) How far will it have gone? A horizontal cable pulls a 200-kg cart along a horizontal track. The tension in the cable is 500 N. Starting from rest, (a) How long will it take the cart to reach a speed of 8.0 m]s? (b) How far will it have gone?

sree lakshmi · Answer

according to work energy theorem change in KE is equal to work done m=200kg ,v=8m/s ,u=0 here work done = force×displacement =F×X =500X change in KE =(1/2)mv 2 = (1/2)200×64 -0 =6400 then 6400 =500X displacement X= 6500/500 = 12.8m now we need to find the time t=? using the equation v=u+at we have f=500=ma =200a then a =5/2 substituting 8=0+(5/2)t t=16/5=3.2sec

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