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Shown below is a 10.0 kg block being pushed by a horizontal force F of magnitude 207.0 N. The coefficient of kinetic friction between the two surfaces is 0.60. Find the acceleration of the block. (Enter the magnitude in m/s2.)

User Francy
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Final answer:

The acceleration of the block is 14.82 m/s^2.

Step-by-step explanation:

To find the acceleration of the block, we need to consider the forces acting on it. The horizontal force applied is 207.0 N. The force of kinetic friction is given by the coefficient of kinetic friction (0.60) multiplied by the normal force (which is equal to the weight of the block, since the block is on a horizontal surface). In this case, the normal force is 10 kg * 9.8 m/s^2 = 98 N. Therefore, the force of kinetic friction is 0.60 * 98 N = 58.8 N. The net force on the block is the applied force minus the force of friction: 207.0 N - 58.8 N = 148.2 N. Finally, the acceleration can be calculated using Newton's second law (F = ma), rearranged to solve for acceleration (a = F/m). Plugging in the values, we get acceleration = 148.2 N / 10.0 kg = 14.82 m/s^2.

User William Roberts
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