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?

Expert's answer

According to the second Newton's law, we have the following equation for the forces:

F = μ·M·g + M·a,

where μ = 0.0610116774619834, M = 47.02 kg, g = 9.8 m/s², a = 19.49 m/s² and F is sought-for force. In fact, the friction force (μ·M·g component) is already given and amounts 28.2 N. Therefore,

F = 28.2[N] + 47.02[kg]·19.49[m/s²] = 944.6198[N].

So, the force needed is

F = 944.6198[N].

F = μ·M·g + M·a,

where μ = 0.0610116774619834, M = 47.02 kg, g = 9.8 m/s², a = 19.49 m/s² and F is sought-for force. In fact, the friction force (μ·M·g component) is already given and amounts 28.2 N. Therefore,

F = 28.2[N] + 47.02[kg]·19.49[m/s²] = 944.6198[N].

So, the force needed is

F = 944.6198[N].

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