problem07_13

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.13: a) The force is applied parallel to the ramp, and hence parallel to the oven’s motion, and so J. 880 m) (8.0 N) 110 ( = = = Fs W b) Because the applied force F is parallel to the ramp, the normal force is just that needed to balance the component of the weight perpendicular to the ramp, , cos α w n = and so the friction force is μ cos k k mg f = and the work done by friction is J, 157 m) 0 . 8 ( 37 cos ) s m (9.80 kg) (10.0 ) 25 . 0 ( cos 2 k f - = ° - = - = s mg W keeping an extra figure. c) J 472 37 sin ) m 0 . 8 )( s m 80 . 9 )( kg 0 . 10 ( sin 2 = ° = mgs , again keeping an extra figure. d) J. 251 J 157 J 472 J 880 = - - e) In the direction up the ramp, the net force is
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Unformatted text preview: N, 31.46 ) 37 cos ) 25 . ( 37 )(sin s m kg)(9.80 . 10 ( N 110 cos sin 2 k = ° + °-=--mg mg F so the acceleration is . s m 3.15 kg) 10.0 N) 46 . 31 ( 2 = The speed after moving up the ramp is , s m 09 . 7 m) (8.0 ) s m 15 . 3 ( 2 2 2 = = = as v and the kinetic energy is J. 252 ) 2 1 ( 2 = mv (In the above, numerical results of specific parts may differ in the third place if extra figures are not kept in the intermediate calculations.)...
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