April 11, 2008
PHY2053Discussion
Quiz 10 (Chapter 9.6 -9.8 & 13.1-13.4)
Name: UFID:
*1. (5 pts)A light spring with spring constant of 200 N/m rests vertically on the bottom of a large container of water(density = 1000 kg/m³). A 5.00-kg block of wood(density = 600 kg/m³) is attached to the upper end of the spring, and the block-spring system is allowed to come to static equilibrium. What is the elongation of the spring?
Three forces are exerted on the block, buoyancy, gravity and spring force. Since the block is in equilibrium, the net force acting on it is zero. Thus we have
B-mg-kΔL = 0 ⇒ΔL = (B-mg)/k
The buoyant force is equal to the weight of displaced water and the volume of the wood is given by the mass divided by the density. Therefore,
ΔL = (ρVg-mg)/k = [(ρ/ρ’)-1]mg/k = [(1000/600)-1]×5×9.8/200 = 0.163 m
**2. (5 pts)A jet of water squirts out horizontally from a hole near the bottom of a large water tankopen to the atmosphere at the top. The tank is placed at the edge of a table, which is 1.00 m high. The jet of water hits the floor 0.800 m away from the bottom of the table. What is the height of the water level in the tank?
First we calculate the velocity of the jet when it leaves the hole:
Δy = -(1/2)gt²⇒t = √(-2y/g) = 0.452 s
Δx = vt ⇒v = Δx/t = 1.77 m/s
Applying Bernoulli’s equation at the surface of the water and at the hole, we have
P₀+ρgh = P₀+(1/2)ρv²⇒h = v²/(2g) = 0.160 m
*3. (5 pts)A 2.00-kg block is attached to a horizontal spring with a force constant of 800 N/m. The spring-block system is on a frictionless surface. The spring is stretched 15.0 cm and released. Find the maximum speed of the block.
The speed of the block is maximum when it passes the equilibrium, thus we have
(1/2)kA² = (1/2)mv²⇒v = √(k/m)A = √(800/2)×0.15 = 3.00 m/s
(We take the positive root because speed is the magnitude of velocity. Always check if you’re using SI units.)
***4. (5 pts)A spring with a spring constant of 400 N/m is situated on a 30.0°frictionless incline. The lower end of the spring is fixed and a 2.00-kg block is attached to the upper end. The block is attached to a string that passes over a frictionless pulley at the top of the incline and then connected to a 3.00-kg hanging object. The system is initially at rest. Now if you cut the string the block starts simple harmonic motion. Find the amplitude of the motion.
Since the system is initially at rest, the initial elongation of the spring is given by
-kL-mgsinθ+T = 0 (Equilibrium condition for the block.)
Mg-T = 0 (Equilibrium condition for the hanging object.)
Mg-mgsinθ-kL = 0 (The sum of the two equation)
⇒L = (Mg-mgsinθ)/k = (3-2sin30˚)×9.8/400 = 0.0490 m
When the string is cut, the block undergoes simple harmonic motion and the center of oscillation is equilibrium position (not unstretched position). The equilibrium position is given by
-mgsinθ-kL₀= 0 ⇒L₀= -mgsinθ/k = -0.0245 m (The negative sign shows the spring is compressed.)
The amplitude is given by
A = L-L₀= 0.049-(-0.0245) = 0.0735 m