## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$F_{max} = 1500~N$
The impulse is equal to the area under the force versus time graph. We can find the area under the graph as: $Area = \frac{1}{2}(F_{max})(8\times 10^{-3}~s)$ $Area = (F_{max})(4\times 10^{-3}~s)$ To find the value of $F_{max}$, we can equate the area under the graph to the impulse. $(F_{max})(4\times 10^{-3}~s) = 6.0~N~s$ $F_{max}= \frac{6.0~N~s}{(4\times 10^{-3}~s)}$ $F_{max} = 1500~N$