
William W. answered 03/04/20
Experienced Tutor and Retired Engineer
For problem 9:
W = F•d
But, the direction of the travel is in the x-direction (the "d" in the equation) so only the portion of the 47N that is being applied in the x-direction contributes to the work being done.
Fx = 47•cos(23°) = 43.264 N
So W = (43.264)(5) = 216.32 Nm (or joules) but since there are only 2 sig figs in the "givens" we would round the answer to 220 Nm (or 220 j)
For problem 10:
This is the same type of problem. Only the portion of the 60N that is applied in the x-direction contributes to the work done in the x-direction. so Fx = 60•cos(25°) = 54.378 N and W = F•d = (54.378)(55) = 2990.82 Nm. Again, we need to round based on the number of sig figs. This problem has only 1 sig fig in the 60 N so our answer needs to have 1 sig fig, so we round to 3000 Nm (or 3000 j)