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trasher [3.6K]
2 years ago
8

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.
Physics
1 answer:
pashok25 [27]2 years ago
6 0

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

∑ <em>F</em> = -<em>f</em> = <em>ma</em> = (4.00 kg) <em>a</em>

Assuming constant acceleration <em>a </em>, the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2<em>a</em> (4.00 m)

Solve for the acceleration :

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

Then the frictional force exerted a magnitude of

-<em>f</em> = (4.00 kg) (-14.8 m/s²)

<em>f</em> ≈ 59.4 N

and was directed opposite the block's motion.

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