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# Answer to Question #21997 in Mechanics | Relativity for tiffany

Question #21997
A concrete block with a mass of 47.02kg is to be pushed along a level floor. The force needed for constant velocity is 28.2 N and the coefficient of kinetic friction is 0.0610116774619834, what force will be needed to accelerate the block at 19.49 m/s/s?
1
2013-01-23T08:32:07-0500
According to the second Newton&#039;s law, we have the following equation for the forces:

F = &mu;&middot;M&middot;g + M&middot;a,

where &mu; = 0.0610116774619834, M = 47.02 kg, g = 9.8 m/s&sup2;, a = 19.49 m/s&sup2; and F is sought-for force. In fact, the friction force (&mu;&middot;M&middot;g component) is already given and amounts 28.2 N. Therefore,

F = 28.2[N] + 47.02[kg]&middot;19.49[m/s&sup2;] = 944.6198[N].

So, the force needed is

F = 944.6198[N].

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