No Se

CHAPTER 3

3.1

Solution:
The work done is equal to the change in kinetic energy 1 U 1−2 = mgx = mv 2 , or 2 v = 2 gx = 14 m/s

5kg 10m

3.2

Solution: Draw an FBD to see the forces acting. The 100 lb weight has a mass of 100/32.2 slugs. Compare the change in kinetic energy from 0 to v ft/s with the work done against the force of friction. Summing forces in the vertical directionyields 100-N + 50 sin 30° = 0, or N = 125 lb. Thus f = µN = (0.2)(125) = 25 lb. The force in the horizontal direction is fx = 50 cos 30° - 25. Thus,

100 50 30°

f N

(43.3 − 25)20 =

1 100 2 v and v = 15.35 ft/s 2 32.2

3.3

Solution:

mg 4m

N

f 45

Let ∆ denote the spring compression. The net force acting along the incline is mg sin 45° - f where f is the friction force.Summing forces perpendicular to the incline yields N = mg cos 45° = (2)(9.81)(0.707). Thus f = µN = (0.6)(2)(9.81)(0.707). Now the net force along the incline will move a distance (4 + ∆)m. The work of the compressed spring is -k∆2/2. Setting the work done moving the mass down the incline and compressing the spring to zero yields: ∆2 o =0 (mg sin 45 − f )(4 + ∆) − k 2 Writing this all in terms of ∆yields the quadratic equation 0.707(2)(9.81)(1-0.6)(4+∆) – 250 ∆2 = 0, or 32.2 + 5.55 ∆ - 250 ∆2 = 0, or ∆ = 0.309 m

3.4

Solution: Assuming no friction, the only forces acting on the particle are gravity and the normal force as indicated in the FBD part of the sketch.

m m θ R

mgsinθ mgcosθ • v N

Set the work done by the weight equal to the change in kinetic energy where v denotes thevelocity as it leaves the surface (normal force does no work). 1 1 2 mg (R − R cosθ ) = mv 2 − mv0 (1) 2 2 Summing the forces in the normal direction of the particle using normal-tangential coordinates yields: ΣFn = ma = − N + mg cosθ. Let β be the angle where N = 0 and using the expression for the normal acceleration yields mv 2 (2) mg cos β = R 2 From (1) with θ = β: v 2 = 2gR(1 − cos β) + v0 .Substitute this into Eq. (2) to get 2 2 ⎛ 2gR + v0 −1 ⎜ 2gR + v 0 ⎞ 2 gR cos β = 2gR(1− cos β ) + v0 or cos β = , thus β = cos ⎝ 3gR ⎠ 3gR

3.5

Solution:
mg θ0 θ R

N

Sketch the geometry of the system. Set the work done equal to the change in kinetic energy to get: U θ − U θ 0 = mgR(sin θ − sin θ 0 ) = Solving for v yields 1 2 1 2 mv − m(0) 2 2

v = 2 gR(sin θ − sin θ 0 ) 3.6Solution:
mg 5kg

Fs

1m ∆

Since the spring is initially compressed in the dimensions given in the drawing, the work ⎛1 2 done corresponding to the spring k (.05) ⎞ must be accounted for in the calculation of ⎝2 ⎠ the total work of the particle. The initial and final kinetic energy is zero, therefore:

500 (∆ + 0.05)2 − (0.05)2 = 0 2 Here ∆ is the additional compression of the spring caused bythe 5 kg mass. Expanding yields: 5 g (1 + ∆) −

[

]

49.1 + 49.1∆ − 250[∆2 + 0.1∆ ]= 0 or 250∆2 − 24.1∆ − 49.1 = 0 The positive solution for ∆ is ∆ = 0.494 m Thus the maximum compression is 0.05 + 0.494 = 0.544 m 3.7 The top spring (k = 200) compresses an amount ∆ while the “bottom” spring (k = 500) compresses an amount (∆ - 0.1). Therefore the total work done is equal to zero:

20 g (∆ )−

1 (200)∆2 − 1 (500)(∆ − 0.1)2 = 0 2 2

− 350∆2 + 246.2∆ − 2.5 = 0 ∆ = 0.693m 3.8 This follows the solution of 3.7 except the force due to gravity acts over the distance (0.2 + ∆) rather than ∆. The total work becomes:
20 g (∆ + 0.2 ) − 1 (200)∆2 − 1 (500)(∆ − 0.1)2 = 0 2 2

− 350∆2 + 246.20 + 36.74 = 0 ∆ = 0.83m 3.9 From the figure, the work done by gravity is mgRsinθ. Let β denote thevalue of θ for which the mass comes to rest so that the arc formed by the compressed spring is defined by β - π/2. The spring is thus compressed R(β - π/2). The total work is:

π⎞2⎤ 1 ⎡ 2⎛ mgRsin β − k ⎢ R ⎝ β − ⎠ ⎥ = 0 2 ⎦ 2 ⎣
This transcendental equation with m = 2 kg, R = 0.3m, k = 200 N/m takes the form

π ⎛ 5.886sin β − 9⎝ β − ⎞ = 0 2⎠
2

Numerical solution yields:

β = 2.276rad =...