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 Joined: 11 Jun 2009 Post: 23
15 Jun 2009 21:07:18 IST
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A truck which has a horizontal surface carries an 80 kg crate. It starts from rest and attains a spe
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A truck which has a horizontal surface carries an 80 kg crate. It starts from rest and attains a speed of 70 km/h in a distance of 70 km on a level road with constant acceleration .Calculate the work done by the frictional force acting on the crate during this interval, given that coefficient of static friction is 0.25 & coefficient of kinetic friction is 0.20

#### Comments (3)

Forum Expert
Joined: 4 Jun 2009
Posts: 72
15 Jun 2009 23:26:53 IST
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in this case the acceleration of the truck comes out to be 0.0027m/s^2.hence there must be a pseudo force acting on the crate which is given by 0.0027*80 = 0.216N.now the limiting friction that can act on the block is given by0.25*(80*9.8)=196 N.. since this value is much greater than the pseudo force , the crate must remain stationery with the static friction force balancing the pseudo force(0.216N).Since static friction does not do any work the work done in this case is 0

Forum Expert
Joined: 4 Jun 2009
Posts: 72
15 Jun 2009 23:28:18 IST
0 people liked this

in this case the acceleration of the truck comes out to be 0.0027m/s^2.hence there must be a pseudo force acting on the crate which is given by 0.0027*80 = 0.216N.now the limiting friction that can act on the block is given by0.25*(80*9.8)=196 N.. since this value is much greater than the pseudo force , the crate must remain stationery with the static friction force balancing the pseudo force(0.216N).Since static friction does not do any work the work done in this case is 0

New kid on the Block

Joined: 11 Jun 2009
Posts: 23
16 Jun 2009 18:48:20 IST
1 people liked this

answer is 8.3 KJ

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