Question

asking for question 2 and part c of 1

Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to report the mass truthfully). (a) What is the force constant of the spring in such a scale if it stretches 6.49 cm for a 11.3 kg load? 1706.317 N/m (b) What is the mass of a fish that stretches the spring 5.50 cm? 9.58 kg (c) How far apart are the half-kilogram marks on the scale? 28.71x How does the distance between the half-kilogram marks compare to the amount a half-kilogram mass stretches the spring? mm Additional Materials Reading 2. -11 points OSColPhys2016 16.2. WA.007.Tutorial. My Notes Ask Your Teacher You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 86.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 6.57 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 2.04 s. Determine the following. (a) force constant of the springs N/m (b) mass of the trailer C kg (c) frequency of the oscillation Hz (d) time it takes for the trailer to bounce up and down 10 times

Answer 1

c)

mg = kx

0.5* 9.8 = 1706.317 x

x = 2.87 mm

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2.

a)

since springs are arranged in parallel so equivalent constant = 2k

2 *k = F/x = 86.2* 9.8 / 0.0657 = 12857.84 N/m

k = 6428.92 N/m

b)

angular frequency

w = 2pi / T = 2pi / 2.04 = 3.08 rad/s

sqrt (k/m) = w

12857.84 / m = 3.08^2

m = 1355.4 Kg

now, M = 1355.4 - 86.2 = 1269.2 kg

c)

f = 1/T = 1/2.04 = 0.49 Hz

d)

t = 10 * 2.04 = 20.4 s

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