An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.

The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is

∑ F = -f = ma = (4.00 kg) a

Assuming constant acceleration a , the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)

Solve for the acceleration :

a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²

Then the frictional force exerted a magnitude of

-f = (4.00 kg) (-14.8 m/s²)

f ≈ 59.4 N

and was directed opposite the block's motion.

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answer: decibel

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