A 3.0-kg block is suspended from a spring of a negligible mass and stretches the

Question

A 3.0-kg block is suspended from a spring of a negligible mass and stretches the spring 0.1 m. (a)What is the force constant of the spring? (b) What is the time period and frequency of oscillation of the system? (c) Find the potential energy at the highest point, (d) What is the kinetic energy at the point of equilibrium? A copper pot with a mass of 0.5 kg contains 0.17 kg of ice at 0^degreeC. A 0.2-kg block of iron at 850^degree C is dropped into the pot. Find the final temperature (C_cu= 390J/kg^degreeC, C_Fe= 470J/kg^degreeC, L_f= 3.34*10^5J/kg, Goo = 4190J/kg^degreeC). Two resistors, A and B, are connected in series to a 6.0-V battery. A voltmeter connected across resistor A measures a voltage of 4.0 V. When the two resistors are connected in parallel across the 6.0-V battery, the current through B is found to be 2.0 A. Find the resistances of A and B. A beam of particles is moving in the electro-magnetic field where the electric component of the field is 4.12 times 10^5 V/m and the magnetic component is 5.71 times 10^-2T, with both fields normal to the beam and to each other. The field produces no deflection of the protons. (a)What is the speed of the proton beam? (b)What is the radius of the proton orbit when the electric field is removed? (c) Through what potential difference would the particle travel to be accelerated to acquire this speed? An electric charge is placed in the magnetic field B. (a) Under what circumstances if any will the force on the charge be zero? (b) Under what circumstances will it be a maximum? Angle theta = 90^degree, (b) theta = 0^degree, (b) velocity = 0 B is perpendicular to v, (b) B is parallel to q Angle theta = 90^degree, (b) angle theta = 180^degree. v = 0, (b) v is perpendicular to B You double your distance from a sound source that is radiating equally in all directions. What happens to the intensity of the sound? It reduces to one-half its original value. one-sixteenth its original value. one-fourth its original value. none of the above.

Explanation / Answer

7. m = 3 kg
x = 0.1 m
a) mg = kx
k = 294 N/m
b) T = 2*pi*sqroot(m/k) = 0.636 s
c) PE at highest point = 3*9.8*0.2 + 0.5*294*0.1^2 = 7.35 J
d) KE at equilibrium = 0.5mv^2 = 0.5kx^2 + mgx
KE = 4.41 J

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