Question
A 42.0 kg swimmer with an initial speed of 1.50 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.00 m, what was the magnitude of the force exerted on her by the water?
Answers
Henry
F = m*a
a = (V^2-Vo^2)/2d
a = (0-(1.5^2))/4 = -0.5625 m/s^2.
F = 42 * (-0.5625) = -23.63 N.
The negative sign means the force opposes the motion.
a = (V^2-Vo^2)/2d
a = (0-(1.5^2))/4 = -0.5625 m/s^2.
F = 42 * (-0.5625) = -23.63 N.
The negative sign means the force opposes the motion.