Answer:
18.83m/s
Step-by-step 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