A.) An 90 kg person steps into a car of mass 2646 kg, causing it to sink 2.35 cm

Question

A.) An 90 kg person steps into a car of mass 2646 kg, causing it to sink 2.35 cm on its springs. Assuming no damping, with what frequency will the car and passenger vibrate on the springs?

B.) A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 19 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)
F2 = ________xF1

Explanation / Answer

apply the relation between Force on spring is given by F = kx

so spring constnt K = F/x

k = mg/x

k = (2464+90)/0.0235

K = 1.08 e 5 N/m

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let me be the mass of tug boat

so formula for force is F = ma

here a = v/t

also the conersion factor is 1 kmph = 5/18 m/s

so a1 = 5*5/18/10 = 0.1388 m/s^2

similarly accleration a2 = 19*5/18/10 = 0.5277 m/s^2

so

Now the ratio of forces F2/F1 = ma2/ma1 =   a2/a1

F2/F1 = 0.5277/0.1388

F2/F1 = 3.808

F2 = 3.8 times of F1

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