Physics Solution Manual for 1100 and 2101

# The work is related to the helicopters kinetic and

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: _____________________ = 61. SSM WWW REASONING After the wheels lock, the only nonconservative force acting on the truck is friction. The work done by this conservative force can be determined from the work-energy theorem. According to Equation 6.8, Wnc = Ef − E0 = ( 1 mv 2 f 2 )( + mghf − 1 mv 2 0 2 + mgh0 ) (1) where Wnc is the work done by friction. According to Equation 6.1, W = ( F cos θ )s ; since the force of kinetic friction points opposite to the displacement, θ = 180° . According to Equation 4.8, the kinetic frictional force has a magnitude of f k = µk FN , where µ k is the coefficient of kinetic friction and FN is the magnitude of the normal force. Thus, Wnc = ( f k cosθ ) s = µ k FN (cos 180°) s = − µ k FN s Since the truck is sliding down an incline, we refer to the free-body diagram in order to determine the magnitude of the normal force FN . The free-body diagram at the right shows the three forces that act on the truck. They are the normal force FN , the force of kinetic friction fk , and the weight mg . The weight has been resolved into its components, and these vectors are shown as dashed arrows. From the free body diagram and the fact that the truck does not accelerate perpendicular to the incline, we can see that the magnitude of the normal force is given by (2) FN fk mg 15° mg cos 15° mg sin 15° FN = mg cos15.0° Therefore, Equation (2) becomes Wnc = − µk ( mg cos15.0° ) s (3) Chapter 6 Problems 323 Combining Equations (1), (2), and (3), we have − µk ( mg cos 15.0° ) s = ( 1 mv 2 f 2 + mghf )−( 1 mv 2 0 2 + mgh0 ) (4) This expression may be solved for s. SOLUTION When the car comes to a stop, vf = 0 m/s. If we take hf = 0 m at ground level, then, from the drawing at the right, we see that h0 = s sin 15.0° . Equation (4) then gives s h0 15° (11.1 m/s )2 s= = = 13.5 m 2 g ( µ k cos15.0° − sin15.0° ) 2 ( 9.80 m/s 2 ) ( 0.750 ) cos15.0° − sin15.0° 2 v0 ______________________________________________________________________________ 62. REASONING The work W done is equal to the average power P multiplied by the time t, or (6.10a) W = Pt The average power is the average power generated per kilogram of body mass multiplied by Armstrong’s mass. The time of the race is the distance s traveled divided by the average speed v , or t = s / v (see Equation 2.1). SOLUTION a. Substituting t = s / v into Equation 6.10a gives 135 ×103 m W s 6 W = Pt = P = 6.50 75.0 kg = 5.48 ×10 J v 12.0 m/s kg 1444 24444 4 3 P ( ) b. Since 1 joule = 2.389 × 10−4 nutritional calories, the work done is 2.389 ×10−4 nutritional calories W = 5.48 ×106 joules = 5.48 ×106 joules 1 joule ( ) = 1.31× 103 nutritional calories ______________________________________________________________________________ 324 WORK AND ENERGY 63. SSM REASONING AND SOLUTION One kilowatt ⋅ hour is the amount of work or energy generated when one kilowatt of power is supplied for a time of one hour. From 3 Equation 6.10a, we know that W = P t . Using the fact that 1 kW = 1.0 × 10 J/s and that 1h = 3600 s, we have 1.0 kWh = (1.0 × 10 3 J/s)(1 h)= (1.0 × 103 J/s)(3600 s)= 3.6 × 10 6 J ______________________________________________________________________________ 64. REASONING The average power is defined as the work divided by the time, Equation 6.10a, so both the work and time must be known. The time is given. The work can be obtained with the aid of the work-energy theorem as formulated in Wnc = ( 1 2 ) 2 mvf2 − 1 mv0 + ( mghf − mgh0 ) (Equation 6.6). Wnc is the work done by the lifting 2 force acting on the helicopter. In using this equation, we note that two types of energy are changing: the kinetic energy ( mv ) and the gravitational potential energy (mgh). 1 2 2 The kinetic energy is increasing, because the speed of the helicopter is increasing. The gravitational potential energy is increasing, because the height of the helicopter is increasing. SOLUTION The average power is P= Wnc t (6.10a) where Wnc is the work done by the nonconservative lifting force and t is the time. The work is related to the helicopter’s kinetic and potential energies by Equation 6.6: Wnc = ( 1 2 ) 2 mvf2 − 1 mv0 + ( mghf − mgh0 ) 2 Thus, the average power is W P = nc = t P= 1 2 ( 1 mv 2 f 2 2 − 1 mv0 2 ) + ( mgh f t − mgh0 ) = 1m 2 (v 2 f 2 − v0 ) + mg ( h f − h0 ) t ( 810 kg ) ( 7.0 m /s )2 − ( 0 m /s )2 + (810 kg ) ( 9.80 m /s2 ) [8.2 m − 0 m] = 2.4 × 104 W 3.5 s ______________________________________________________________________________ Chapter 6 Problems 325 65. REASONING The average power is given by Equation 6.10b ( P = Change in energy ). The time Time is given. Since the road is level, there is no change in the gravitational potential energy, and the change in energy refers only to the kinetic energy. According to Equation 6.2, the kinetic energy is 1 mv 2 . The speed v is given, but the mass m is not. However, we can obtain 2 the mass from the given weights, since the weight is mg. SOLUTION a. Using Equations 6.10b and 6.2, we find that the average power is 2 2 Change in energy KE f − KE 0 1 mvf − 1 mv0 2 2 P= = = Time Time Time Since the car starts from rest, v0 = 0 m/s, and si...
View Full Document

## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online