A brick with mass m_1 = 2.5 kg can move without friction on a horizontal plane.
ID: 1469047 • Letter: A
Question
A brick with mass m_1 = 2.5 kg can move without friction on a horizontal plane. The brick is attached to a wall by an ideal spring with spring constant k = 2300 N/m.A second brick with mass m_2 = 1.2 kg lies on top of the brick as shown in Fig. 4. The static coefficient of friction between the bricks is mu_s = 0.56. Assuming that m_2 does not slide on m_1 calculate the period, T, of small angle oscillations for this system. Calculate the maximum amplitude of oscillation at which m_2 can follow along without beginning to slide on m_1 A grandfather clock is based on the period of oscillation of a pendulum. The clock keeps perfect time on the first floor of the Sears tower but when carefully moved to the top floor h = 436 m above the surface of the earth, the clock is found to be slow. What is the reason for this? How many seconds docs the clock loose in one day on the top floor? (The radius of the earth is R-6.37 10^6 m) A water faucet in a house is h = 8 m below the water surface in the local water tower. The radius of the orifice in the fully opened faucet is r=0.01 m. We assume non-viscous flow in this problem. How long does it take to fill a 20 liter bucket with water using this faucet.Explanation / Answer
Problem - 2
(a)
Period of Oscillation, T = 2(m/k)
Where,
m = 2.5 + 1.2 Kg
k = 2300 N/m
T = 2*(3.7/2300)
T = 0.252 s
(B)
Maximum Force for which m2 will not slide on m1 = Friction Force between them
Fr = u * m2*g
Force = Mass * Acceleration , Therefore
m * a = 0.56 * m * 9.8 N
a = 5.488 N
Now, Force due to Spring, F = k*x
Force = Mass * Acceleration, Therefore
(m1 + m2) * a = 2300 * x
x = ((m1 + m2) * a)/ 2300
x = ((1.2 + 2.5) * 5.488) / 2300
x = 0.00883 m
x = 0.883 cm
Maximum Amplitude of oscillation, x = 0.883 cm
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