You are here
A 2 kg object moving at 10 m s
−1
enters a rough patch where friction is 7 N, plus an extra 3 N opposing force. How far does it travel before stopping?
View Answer
Total opposing force = 7 + 3 = 10 N. Acceleration \( a = \dfrac{F}{m} = \dfrac{-10}{2} = -5\ \text{m s}^{-2} \).
Using \( v^2 = u^2 + 2as \) with v = 0, u = 10: \( 0 = 100 + 2(-5)s \), so \( s = \dfrac{100}{10} = \mathbf{10\ m} \).
← Back to concept: 6.5 Newton's Second Law