static and kinetic coefficient of friction

Could u guys help me out of this? I am confusing about this exercise because it seems to be 2 systems of forces in one. ( I may get it wrong).
Thanks

A flatbed truck, which carries an 80 kg crate, starts form rest and attains a speed of 72 km/h in a distance of 75 m on a level road with constant acceleration. Calculate the work done by the friction force acting on the crate during this interval if the static and kinetic coefficients of friction between the crate and the truckbed are 0.3 and 0.28 respectively.

Reply 1

Jonny W

The crate does not slip if the truck's acceleration a satisfies

80a <= 0.3(normal force between truck and crate)
ie, 80a <= 0.3*80g
ie, a <= 0.3g
ie, a <= 2.94

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Final speed of truck = 72km/h = 72000m/h = 20m/s
Average speed of truck = 10m/s
Time taken for truck to cover 75m = 7.5s
a = 20/7.5 = 8/3

--

The crate does not slip because 8/3 <= 2.94.

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Final speed of crate = Final speed of truck = 20m/s
Work done by frictional force = Final KE of crate = (1/2)(80)(20^2) = 16000J

Reply 2

duchiep

This is really a trick question, it took me 4 h to think about it. !!!!!!!!!!
We might not need the kinetic coefficient of friction in this part but in the next part they mention about the case when the static coefficient is 0.26 and the kinetic one is 0.25.
How can I work out the work done by friction force in this case ??

PS: I got this quest from my teacher he said he took it from Practice in Physics book.

Reply 3

Jonny W

duchiep

We might not need the kinetic coefficient of friction in this part but in the next part they mention about the case when the static coefficient is 0.26 and the kinetic one is 0.25.
How can I work out the work done by friction force in this case ??
This time (big surprise) the crate does slip because 8/3 > 0.26g.

Let b be the acceleration of the crate. Let N and F be the normal and frictional forces between the crate and truck.

Then

N = 80g
F = 0.25N = 20g
b = F/80 = g/4

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Final speed of crate = 7.5b = (15/8)g
Work done by frictional force = Final KE of crate = (1/2)(80)[(15/8)g]^2 = 13500J