00:01
Here we can say for part a the work done f work sorry the work done w will equal the force times the distance this would be equal to 35 .0 newtons multiplied by 3 .00 meters and this is equaling 105 joules for part b we know that here the total amount of energy that has gone to thermal forms we're going to use equation 831 and you equation 6 -2.
00:36
And we can say that here the change in thermal energy would be equal to the coefficient of kinetic friction times m gd.
00:44
This is equaling 0 .6 .00 multiplied by the mass of 4 .00 kilograms multiplied by 9 .8 meters per second squared multiplied by the same distance of 3 .00 meters and this is equaling 70 .6 joules.
01:05
Now, we know that if 40 joules have gone to the block, this means that here, 70 .6 minus 40 .0 joules, we can say that this is equaling 30 .6 joules has, we can say, has been transferred to the floor, given that the block gained 40 joules of thermal energy.
01:39
We can then say for part c, much of the work 105 joules has been wasted due to the 70 .6 joules of thermal energy generated.
01:50
But there still remains, we can say remaining would be 105.
02:00
The answer in part a, minus the answer to part, rather the calculation we had did for the change in.
02:06
We had done for the change in thermal energy, 70 .6...