A 270 g block connected to a light spring with a force constant of k = 4 N/m osc

Question

A 270 g block connected to a light spring with a force constant of k = 4 N/m oscillates on a horizontal, frictionless surface. The block is released from rest with an initial displacement xi = 5 cm from equilibrium. (A) What is the maximum magnitude of the force exerted by the spring during this motion? (B) At what value of x is the force from the spring equal to P = 40% of the maximum force? (C) At what value of x is the speed of the block equal to 40% of its maximum speed? Conceptualize The block is pulled back to stretch the spring, and is then released. As the block accelerates back and forth, the spring alternatively compresses and stretches, with maximum deformation at the locations where the block momentarily stops to reverse direction, and maximum speed where the block passes through the equilibrium point x = 0. Somewhere between the extremes are locations where the magnitude of the force has the specified value and other locations where the speed attains its specified value. Categorize The problem involves analyzing how the speed of an object and the force of the spring on it vary with position in simple harmonic motion. Analyze (A) What is the maximum magnitude of the force exerted by the spring during this motion? The force is proportional to the displacement, and at the initial displacement produces the maximum force, having a magnitude equal to: | F | = (4 N/m)(5.00 10-2 m) | - kxmax | = (4 N/m)(5.00 10-2 m) = N (B) At what value of x is the force from the spring equal to 40% of the maximum force? Set | F | = | -kx | equal to 40% of the maximum value, without keeping track of the minus sign because the question asks about the magnitude of the force: | -kx | = 40 100 =| kxmax | From this it is seen that the force is reduced to 40% of its maximum magnitude where |x| has 40% of its maximum value, or at x = ± cm. (C) At what value of x is the speed of the block equal to 40% of its maximum speed? We can answer this question from the solutions for the position and velocity as functions of time. Because the block is at its maximum displacement x = x0 at time t = 0, the equation for x is: (1) x = x0 cos(t) with = 0. Therefore the equation for the velocity is: (2) v = -x0 sin(t) whose maximum magnitude is x0. The time t when the speed is 40% of this maximum value therefore satisfies: (3) 40 100 x0 = x0 |sin(t)| This is satisfied for: sin[t] = ± 40 100 But because sin2(t) + cos2(t) =1, Equation (4) reduces to: cos(omegat) = sqrt(1 - sin^2 (omegat)) = +/- sqrt(1 - ((40/(100)))^2) And therefore the locations were the speed has the required values are: x = x_0 cos(omegat) = +/- x_0 sqrt(1 - ((40/(100)))^2)

Explanation / Answer

a)

spring constant is k = 4 N/m

initial speed is Vo = 0 m/sec

initial dispalcement is xi = 5 cm

maximum force is Fmax = k*xi = 4*5*10^-2 = 0.2 N

b) Given that spring force = 0.4*Fmax = 0.4*0.2

F = 0.08

k*x = 0.08

x = 0.08/k = 0.08/4 = 0.02 m = 2 cm

c) Vmax = A*w

w = sqrt(k/m) = sqrt(4/0.27) = 3.85 rad/sec

Vmax = A*w = 0.05*3.85 = 0.1925 m/sec

V = 0.4*Vmax = 0.4*0.1925 = 0.077 m/sec

then using law of conservation of energy

0.5*m*Vmax^2 = (0.5*k*x^2)+(0.5*m*v^2)

0.5*0.27*0.1925^2 = (0.5*4*x^2)+(0.5*0.27*0.077^2)

x = 0.046 m = 4.6 cm

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