001) The force required to stretch a Hooke’s-law spring varies from 0 N to 43 N as we stretch the spring by moving one end 10 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m
002 (part 2 of 2) Find the work done in stretching the spring. Answer in units of J
003) It takes 5.86 J of work to stretch a Hooke’s-law spring 11.2 cm from its unstressed length. How much the extra work is required to stretch it an additional 5.2 cm? Answer in units of J
004) When you try to stretch a bungee cord a distance x, it resists with an opposing force of the form b x2, where b is a constant. If b is measured to be 6 N/m2, how much work does it take to stretch the bungee cord a distance of 1 meter? 1. 2 Joules 2. 3 Joules 3. 0.5 Joule 4. 1 Joule
005 (part 1 of 3) A spring has a force constant of 542.9 N/m. Find the potential energy stored in the spring when the spring is
a) stretched 4.11 cm from equilibrium. Answer in units of J b) compressed 2.52 cm from equilibrium. Answer in units of J
006 (part 2 of 3) b) compressed 2.52 cm from equilibrium. Answer in units of J
007 (part 3 of 3) c) unstretched. Answer in units of J
008) The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the spring constant is 53.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m? Answer in units of J
009) A spring with a force constant of 5.0 N/m has a relaxed length of 2.64 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.65 m. Calculate the elastic potential energy stored in the spring. Answer in units of J
010) What is the acceleration of a 63 kg block of cement when pulled sideways with a net force of 464 N? Answer in units of m/s^2
+33661 Hookes law says force is proportional to extension. F = k*x
1) Extension = 10cm (= 0.1m), force = 43N, so k = 43/0.1 N/m or k = 430N/m
2) Work done is the integral of force *distance with respect to distance
$$work=\int_0^Xkxdx=\frac{1}{2}kX^2$$
work = (1/2)*430*0.12J or work = 21500 J
Most of the others can be answered using a similar approach with Hookes law. The exception is the last one:
10) Force = mass*acceleration
464 = 63*a
a = 464/63 m/s2
a ≈ 7.365 m/s2
.
+33661 Hookes law says force is proportional to extension. F = k*x
1) Extension = 10cm (= 0.1m), force = 43N, so k = 43/0.1 N/m or k = 430N/m
2) Work done is the integral of force *distance with respect to distance
$$work=\int_0^Xkxdx=\frac{1}{2}kX^2$$
work = (1/2)*430*0.12J or work = 21500 J
Most of the others can be answered using a similar approach with Hookes law. The exception is the last one:
10) Force = mass*acceleration
464 = 63*a
a = 464/63 m/s2
a ≈ 7.365 m/s2
.
