Chapter 10 Quiz

Melissa's favorite exercise equipment at the gym consists of various springs. In one exercise, she pulls a handle grip attached to the free end of a spring to 0.80 m from its unstrained position. The other end of the spring (spring constant = 45 N/m) is held in place by the equipment frame. What is the magnitude of the force that Melissa is applying to the handle grip?

(F = kx = 45 x 0.80 =) 36 N

Which one of the following statements is true concerning an object executing simple harmonic motion? A) The object's velocity is never zero. B) The object's acceleration is never zero. C) The object's velocity and acceleration are simultaneously zero. D) The object's velocity is zero when its acceleration is a maximum. E) The object's maximum acceleration is equal to its maximum velocity.

D

A 0.2 kg block is held in place by a force F that results in a 0.10 m compression of a spring beneath the block. The spring constant is 1.0 × 102 N/m. Assuming the mass of the spring is negligible compared to that of the block, to what maximum height would the block rise if the force F were removed.

(Elastic potential energy of spring = 1/2kx2 = 0.5 x 1.0 x 102 x 0.1 x 0.1 = 0.5 J At the maximum height of the block, elastic potential energy of spring = 0 and gravitational potential energy of block = mgh = 0.2 x 9.8 x h = 1.96h J Using conservation of energy, 1.96h = 0.5 h = 0.5/1.96 =) 0.26 m

A ping-pong ball weighs 0.025 N. The ball is placed inside a cup that sits on top of a vertical spring. If the spring is compressed 0.055 m and released, the maximum height above the compressed position that the ball reaches is 2.84 m. Neglect air resistance and determine the spring constant.

(Elastic potential energy of spring = 1/2kx2 = 0.5 x k x 0.055 x 0.055 = 0.0015 k J At the maximum height of the block, elastic potential energy of spring = 0 and gravitational potential energy of ping-pong ball = mgh = 0.025 x 2.84 = 0.071 J Using conservation of energy, 0.0015 k = 0.071 h = 0.071 /0.0015 =) 47 N/m

A spring with a spring constant k = 1,600 N/m is at rest on the bottom of an inclined plane (55). A 7.0 kg block slides down the plane and makes contact with the spring at point A as shown. After contact, the spring is compressed to point B, 0.20 m from point A, where the speed of the block is zero m/s. What was the speed of the block just before contact with the spring?

(Elastic potential energy of spring when block is stationary = 1/2kx2 = 0.5 x 1,600 x 0.20 x 0.20 = 32 J When the block just makes contact with the spring, elastic potential energy of spring = 0 Kinetic energy of block = 1/2mv2 = 0.50 x 7.0 x v2 = 3.5 v2 J During the movement of the spring, gravitational potential energy lost by the block = mgh = 7.0 x 9.8 x 0.20 sin 55o = 11.2 J Using conservation of energy, 3.5 v2 = 32 + 11.2 = 43.2 v2 = 43.2/3.5 = 12.34 v = √12.34 =) 3.5 m/s or 2.4 m/s

A cable stretches by an amount d as it supports a crate of mass M. The cable is cut in half. What is the mass of the load that can be supported by either half of the cable if the cable stretches by an amount d?

(For the same stress as on the full cable, the new cable would stretch by only d/2 Since it has stretched by d, the stress must now be two times the original Therefore the mass must also be two times the original, i.e.) 2M

A ball hung from a vertical spring oscillates in simple harmonic motion with an angular frequency of 2.6 rad/s and an amplitude of 0.075 m. What is the maximum acceleration of the ball?

(Maximum acceleration = Aω2 = 0.075 x 2.6 x 2.6 =) 0.51 m/s

A 10 kg box is at rest at the end of an unstretched spring with constant k = 4,000 N/m. The mass is struck with a hammer giving it a velocity of 6.0 m/s to the right across a frictionless surface. What is the amplitude of the resulting oscillation of this system?

(Maximum elastic potential energy = 1/2kx2 = 0.5 x 4,000 x2 = 2,000 x2 J Maximum kinetic energy of box = 1/2 mv2 = 0.5 x 10 x 6.0 x 6.0 = 180 J Using conservation of momentum, 2,000 x2 = 180 x2 = 180/2,000 = 0.09 x = √0.09 =) 0.3 m

A spring required a force of 5.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m?

(Since F = kx k = F/x = 5.0/0.1 = 50 N/m Work required to stretch the spring = elastic potential energy = 1/2kx2 = 0.5 x 50 x 0.4 x 0.4 =) 4 J

A block is suspended from the ceiling by a long, thin strip of tungsten metal. The strip behaves as a spring. To produce a 0.25 m horizontal deflection of the block, a force of 6.5 N is required. Calculate the spring constant for the tungsten strip.

(Since F = kx k = F/x = 6.5/0.25 =) 26 N/m

A cable stretches by an amount d when it supports a crate of mass M. The cable is replaced by another cable of the same material having the same length and twice the diameter. What is the mass of the load that can be supported by the thicker cable if it stretches by an amount d?

(Since cross sectional area of cable = πr2 The cross sectional area of cable 2/cross sectional area of cable 1 = (2r)2/r2 = 4 Since stress = force/cross sectional area New stress = 1/4 old stress Therefore new strain = 1/4 old strain New cable will stretch by d/4 Since the new cable stretches by d, it will require four times the original stress Therefore mass of new load =) 4M

The acceleration of a certain simple harmonic oscillator is given by a = -(15.8 m/s2) cos (2.51t). What is the amplitude of the simple harmonic motion?

(Since equation for acceleration, a = -Aω2 cos ωt) (In the equation given, Aω2 = 15.8 and ω = 2.51) (A = 15.8/ ω2 = 15.8/(2.51)2 =) 2.51 m

The position of a simple harmonic oscillator is given by x(t) = (0.50)cos(/3t) where t is in seconds. What is the maximum velocity of this oscillator?

(Since equation for displacement, x = A cos ωt In the equation given, A = 0.5 m and ω = π/3 Maximum velocity, v = Aω = 0.50 x π/3 =) 0.52 m/s

Young's modulus of nylon is 3.7 × 109 N/m2. A force of 6.0 × 105 N is applied to a 1.5 m length of nylon of cross sectional area 0.25 m2. By what amount does the nylon stretch?

(Stress = force/area = 6.0 x 105/0.25 = 2.4 x 106 Pas Young's modulus = stress/strain 3.7 x 109 = 2.4 x 106/strain Strain = 2.4 x 106 /3.7 x 109 = 6.5 x 10-4 = Δl/1.5 Δl = 6.5 x 10-4 x 1.5 =) 9.8 x 10^-4 m

The shear modulus of steel is 8.1 × 1010 N/m2. A steel nail of radius 7.5 × 10-4 m projects 0.040 m horizontally outward from a wall. A man hangs a wet raincoat of weight 28.5 N from the end of the nail. Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

(Stress on nail = Force/cross sectional area = 28.5/[π x (7.5 x 10-4)2] = 28.5/(π x 5.625 x 10-7) = 1.6 x 107 N/m2 Strain on nail = x/0.040 Stress/strain = modulus 1.6 x 107/( x/0.040) = 8.1 x 1010 x = (1.6 x 107 x 0.040)/ 8.1 x 1010 =) 7.9 x 10^-6 m

A 5.0 × 10^2 N object is hung from the end of a wire of cross-sectional area 0.010 cm2. The wire stretches from its original length of 200.00 cm to 200.50 cm. Determine the Young's modulus of the wire.

(Stress on wire = force/cross sectional area = (5.0 x 102)/[0.010 x (10-2)2] ) (= (5.0 x 102)/(1.0 x 10-6) = 5.0 x 108 N/m2 ) (Elongation strain on wire = Increase in length/original length =) ((200.50 - 200.00)/200.00 = 0.50/200.00 = 2.5 x 10-3) (Young's modulus = stress/strain = 5.0 x 108/2.5 x 10-3 =) 2.5 x 10^11

A relaxed spring protrudes from an opening 0.0500 meters as shown in Figure A. A 1.00 kg block is found to just force the spring completely into the opening as shown in Figure B. How much potential energy is stored in the spring in Figure B?

(Use equation F = kx k = F/x Force due to block = 1.0 x 9.8 = 9.8 N k = 9.8/0.0500 = 196 N/m Potential energy stored in the spring = 1/2kx2 = 0.5 x 196 x 0.050 x 0.050 =) 0.245 J

In a certain clock, a pendulum of length L1 has a period T1 = 0.95 s. The length of the pendulum is adjusted to a new value L2 such that T2 = 1.0 s. What is the ratio L2/L1?

(Use the equation T = 2π √ (L/g) For pendulum 1, 0.95 = 2π √(L1/9.8) (1) For pendulum 2, 1.0 = 2π √(L2/9.8) (2) Divide equation (2) by equation (1) 1.0/0.95 = [2π √(L2/9.8)]/ [2π √(L1/9.8)] = √(L2/ L1) L2/ L1 = (1.0/0.95)2 =) 1.1

A plastic box has an initial volume of 3.50 m3. It is then submerged below the surface of a liquid and its volume decreases to 3.48 m3. What is the volume strain on the box?

(ΔV = 3.50 - 3.48 = 0.02 m3 Volume strain = 0.02/3.50 =) 0.0057

A simple pendulum consists of a ball of mass m suspended from the ceiling using a string of length L. The ball is displaced from its equilibrium position by a small angle . What is the magnitude of the restoring force that moves the ball toward its equilibrium position and produces simple harmonic motion?

mg(sin @)