(b) Use Newton's second law. The net forces on block <em>M</em> are
• ∑ <em>F</em> (horizontal) = <em>T</em> - <em>f</em> = <em>Ma</em> … … … [1]
• ∑ <em>F</em> (vertical) = <em>n</em> - <em>Mg</em> = 0 … … … [2]
where <em>T</em> is the magnitude of the tension, <em>f</em> is the mag. of kinetic friction between block <em>M</em> and the table, <em>a</em> is the acceleration of block <em>M</em> (but since both blocks are moving together, the smaller block <em>m</em> also shares this acceleration), and <em>n</em> is the mag. of the normal force between the block and the table.
Right away, we see <em>n</em> = <em>Mg</em>, and so <em>f</em> = <em>µn</em> = 0.2<em>Mg</em>.
The net force on block <em>m</em> is
• ∑ <em>F</em> = <em>mg</em> - <em>T</em> = <em>ma</em> … … … [3]
<em />
You can eliminate <em>T</em> and solve for <em>a</em> by adding [1] to [3] :
(<em>T</em> - 0.2<em>Mg</em>) + (<em>mg</em> - <em>T </em>) = <em>Ma</em> + <em>ma</em>
(<em>m</em> - 0.2<em>M</em>) <em>g</em> = (<em>M</em> + <em>m</em>) <em>a</em>
<em>a</em> = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)
<em>a</em> = 1.96 m/s²
We can get the tension from [3] :
<em>T</em> = <em>m</em> (<em>g</em> - <em>a</em>)
<em>T</em> = (10 kg) (9.8 m/s² - 1.96 m/s²)
<em>T</em> = 78.4 N
(c/d) No time duration seems to be specified, so I'll just assume some time <em>t</em> before block <em>M</em> reaches the edge of the table (whatever that time might be), after which either block would move the same distance of
1/2 (1.96 m/s²) <em>t</em>
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(e) Assuming block <em>M</em> starts from rest, its velocity at time <em>t</em> is
(1.96 m/s²) <em>t</em>
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(f) After <em>t</em> = 1 s, block <em>M</em> reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have
∑ <em>F</em> = -<em>f</em> = <em>Ma</em>
The effect of friction is constant, so that <em>f</em> = 0.2<em>Mg</em> as before, and
-0.2<em>Mg</em> = <em>Ma</em>
<em>a</em> = -0.2<em>g</em>
<em>a</em> = -1.96 m/s²
Then block <em>M</em> slides a distance <em>x</em> such that
0² - (1.96 m/s²) = 2 (-1.96 m/s²) <em>x</em>
<em>x</em> = (1.96 m/s²) / (2 (1.96 m/s²))
<em>x</em> = 0.5 m
(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block <em>M</em> and pulley" …)
Meanwhile, block <em>m</em> would be in free fall, so after 1 s it would fall a distance
<em>x</em> = 1/2 (-9.8 m/s²) (1 s)
<em>x</em> = 4.9 m
