The relationship between the extension of an elastic spring x and the force h it
ID: 1501745 • Letter: T
Question
The relationship between the extension of an elastic spring x and the force h it exerts on a mass hanging from it is given by a F=-kx^2 b. F = k/x C. F = - kx d. F= 1/2k/x e. F= -x/k 255 g mass is hooked up to a spring (k = 175 N/m) and moves back and forth on your basic frictionless surface. If the mass is released from rest at x - 0.200 m. (a) find the force acting on the mass, (b) the max acceleration, (c) it's acceleration at x = 0 m. (d) its energy, and (e) its period. Period is defined as (A) The distance traveled in one oscillation (B) The time it takes to travel one oscillation (C) The average velocity over one oscillation (D)The maximum displacement of an oscillation A 355 g mass is attached to a spring (k = 435 N/m). If the system is allowed to oscillate on a frictionless surface, what is the period and frequency of the motion? When 3.0 kg of water is warmed from 10.0 degree C to 80.0 degree C, how much heat energy is needed? Calculate the temperature change when 10,000.0 g of water loses 232,000 J of heat 16 1960 J of heat are added to 500. g of copper taking its temperature from 25.0 degree C to 35.1 degree C. What is the specific heat capacity of copper? Bernoulli's Principle is a statement of (A) energy conservation in dynamic fluids. (B) momentum conservation in dynamic fluids. (C) hydrostatic equilibrium. (D) thermal equilibrium in fluids. (E) mechanical equilibrium in fluids.Explanation / Answer
10. F = - kx
Ans(x)
11.
a) F = kx = 175 x 0.2 =35 N
b) Max acc. = Fmax / m = 35 / 0.255 = 137.25 m/s^2
c) at x = 0 , F = 0
hence a = 0
d) energy = kx^2 /2 = 175 x 0.2^2 /2 = 3.5 J
e) T = 2 pi sqrt(m/k) = 2 pi sqrt(0.255 / 175) = 0.24 sec