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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?

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? 

Grade:12

1 Answers

sree lakshmi
73 Points
one year ago
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)mv2 = (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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