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Unformatted text preview: h ic e8 e k# 4 i b iC a aha 4 Problem 10 ( a ): The acceleration is found by x = arrownortheast v t + 1 2 a ( t ) 2 a = 2 x ( t ) 2 = 0 . 88 m / s 2 . The force is F = 80 . 0 N, so the mass is given by F = ma m = F a = 80 . 0 N . 88 m / s 2 = 90 . 9 kg . ( b ): The final velocity is v f = arrownortheast v + a t = (0 . 88 m / s 2 )(5 . 00 s) = 4 . 4 m / s , so in the following 5.00 s the block moves an additional (4 . 4 m / s)(5 . 00 s) = 22 m . 4 Problem 16 ( a ): The electron starts at rest and travels a known distance, ending with a known velocity. Thus we use v 2 f v 2 = 2 a x a = v 2 f v 2 2 x = (3 . 00 10 6 m / s) 2 2(0 . 0180 m) = 2 . 54 10 14 m / s 2 . ( b ): To find the time to reach the grid we use x = 1 2 a ( t ) 2 t = radicalbigg 2 x a = radicalBigg 2(0 . 0180 m) 2 . 54 10 14 m / s = 1 . 2 10 8 s . ( c ): The force is F = ma = (9 . 11 10 31 kg)(2 . 50 10 14 m / s 2 ) = 2 . 28 10 16 N . 4 Problem 40: The cars initial speed is v = 45 . 0 km hr 1000 m 1 km 1 hr 3600 s = 12 . 5 m / s . The final velocity is v f = 0, and the distance is x = 1 . 8 cm = 0 . 018 m, so v 2 f v 2 = 2 a x a = v 2 f v 2 2 x = (12 . 5 m / s) 2 2(0 . 018 m) = 4300 m / s 2 . This would require a force of F = ma = (850 kg)( 4300 m / s 2 ) = 3 . 7 10 6 N . 4 Problem 43 ( a ): The two crates are tied together, so as one moves, so does the other. Thus the acceleration of the 4.00 kg crate is also 2 . 50 m / s 2 . 1 ( b ): The free body diagram looks like this: T 4.00 kg Newtons second law for this crate says that T = m 1 a = (4 . 00 kg)(2 . 50 m / s 2 ) = 10 . 0 N . ( c ): The free body diagram looks like this: 6.00 kg F T The direction of the net force must be to the right, as this crate is accelerating to the right. Thus we must have F > T ....
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 Summer '08
 Chamberlin
 Acceleration, Force, Mass, Work

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