# Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems: 2-115E

a)$W_{f, man} = 2000 \ lbf.ft = 2.57BTU$ b)$W_{f, block} = -2000 \ lbf.ft = -2.57BTU$

#### Work Step by Step

The total work of the man-block system is zero, which means that the work done by the man must be equal to the work done on the block, in magnitude. I) For the block: Since the plane is horizontal: $F_{W} = F_{N}$ The equation for the friction work is : $W_f = \int F_f dx$ $W_f = \mu F_N\Delta x$ With $\mu = 0.2$ $F_N = 100\ lbf$ $\Delta x = 100\ ft$ $W_f = 2000 \ lbf.ft$ $W_f = 2000 \ lbf.ft \times \frac{1\ BTU}{778.2\ lbf.ft} = 2.57 BTU$ Using the formal sign convention: $W_{f, block} = -2000 \ lbf.ft = -2.57BTU$ $W_{f, man} = 2000 \ lbf.ft = 2.57BTU$

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