(1) W  =  m g h  =  75 kg×9.8 m/s²×10 m  =  7,350 J
(3) F  =  m n  =  m m g, W  =  F d  =  0.7×150 kg ×9.8 m/s² ×12.3 m  =  1.3×10⁴ J
(5) D PE  =  W  = m g h ® h  =  W/(mg)  = 115 J/(0.325 kg×9.8 m/s²)  =  36.1 m
(7) Must overcome both the component of the gravitational force down the plane, m g sinq and the friction force when present, m m g cosq. (a) Ignoring friction, the force is F  =  1000 kg×9.8 m/s²×sin(17.5^°)  =  2,950 N. Thus the work done is F×d  =  2950 N×300 m  =  8.84×10⁵ J.
(b) With friction, F  =  2950 N + 0.25×1000 kg×9.8 m/s²×cos(17.5^°)  =  5.29×10³ J. So the work is W  =  5.29×10³×300 m  =  1.59×10⁶ J
(17) mv²/2  =  KE, so v  =  (2×KE/m)^1/2  =  (2×6.21×10^-21J/5.31×10^-26kg)^1/2  =  480 m/s
(19) |W|  =  KE  =  m v²/2  =  9.11×10^-31kg×(1.9×10⁶ m/s²)²/2  =  1.64×10^-18 J
Since the force is directed opposite the velocity (deceleration), the sign is negative.
(21) % increase   =  ([(100/90)² - 1]/90)×100  =  23%
(25) m m g d  =  m v²/2 ® v  =  (2m g d)^1/2  =  (2×0.42×9.8 m/s²×88 m)^1/2  =  27 m/s. Note that the mass is common to both sides of the equation so cancels.
(29) k x²/2  =  PE ® x  =  (2 PE/k)^1/2  =  (2×25 J/440 N/m)^1/2  =  0.34 m
(31) D PE  =  m g h  = 64 kg×9.8 m/s²×4 m  =  2500 J
(37) mv²/2  =  m g h ® v  =  (2 m g h)^1/2  =  (2×9.8×1.35)^1/2m/s  =  5.14 m/s. Note that the incline angle is not needed to solve this problem.
(45) E  =  PE₀  =  k x₀²/2  =  KE + PE  =  mv²/2 + k x²/2. Note: the book answer involving a gravitational potential m g x is in error.
(49) 95 km/h  =  26.39 m/s. Thermal energy converted is E  =  2×mv²/2  =  m v²  =  6500 kg×(26.39 m/s)²  =  4.53×10⁶ J
(59) P  =  F×v  =  18 hp×746 W/hp  =  1.34×10⁴ J/s.
90 km/h  =  25 m/s ® F  =  (1.34×10⁴ J/s)/25 m/s  =  540 N
(63) P  =  W/t ®  W  =  P×t  =  3 hp×746 W/hp×1 h×3600 s/h  =  8.1×10⁶ J

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File translated from T_EX by T_TH, version 1.95.
On 4 Oct 1999, 11:26.