A 24.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?

18.83m/s

Explanation:

In the first 11meters, we can calculate the Kinectic energy since we know that

the floor is frictionless then work done by the horizontal force= Kinectic energy gain by the block in the first 11m.

Then Kinetic Energy = wordone

But work done= Force × distance

Kinetic Energy=(225×11)= 2475J

In the next 10.0 m the coefficient of friction is 0.20, the Kinectic energy has equal value to the difference in workdone by both the horizontal and frictional force

K.E= (Force× distance) - ( mass × gravity× coefficient of friction × distance at 10 m)

K.E= [(225×10)-(.20× 9.8×10× 24)]

K.E= 2250-470.4

= 1779.6J

The addition of the Kinectic energy above give us the total Kinectic energy experience by the crate.

Total Kinectic energy= 1779.6+2475

= 4254.6J

the final speed of the crate after being pulled these 21.0 m?

Total distance= 11m + 10m = 21 m

Then the final speed can be calculated from the total Kinectic energy, since we know that

K.E= 0.5mv^2

V= √(2K.E/m)

= √(2×4254.6)/24

Final speed v = √354.55

Final speed v= 18.83m/s

Therefore, the final speed of the crate after being pulled these 21.0 m is 18.83m/s

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