A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?

(b) Use Newton's second law. The net forces on block M are

• ∑ F (horizontal) = T - f = Ma … … … [1]

• ∑ F (vertical) = n - Mg = 0 … … … [2]

where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.

Right away, we see n = Mg, and so f = µn = 0.2Mg.

The net force on block m is

• ∑ F = mg - T = ma … … … [3]

You can eliminate T and solve for a by adding [1] to [3] :

(T - 0.2Mg) + (mg - T ) = Ma + ma

(m - 0.2M) g = (M + m) a

a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)

a = 1.96 m/s²

We can get the tension from [3] :

T = m (g - a)

T = (10 kg) (9.8 m/s² - 1.96 m/s²)

T = 78.4 N

(c/d) No time duration seems to be specified, so I'll just assume some time t before block M 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) t

(e) Assuming block M starts from rest, its velocity at time t is

(1.96 m/s²) t

(f) After t = 1 s, block M 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

∑ F = -f = Ma

The effect of friction is constant, so that f = 0.2Mg as before, and

-0.2Mg = Ma

a = -0.2g

a = -1.96 m/s²

Then block M slides a distance x such that

0² - (1.96 m/s²) = 2 (-1.96 m/s²) x

x = (1.96 m/s²) /  (2 (1.96 m/s²))

x = 0.5 m

(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)

Meanwhile, block m would be in free fall, so after 1 s it would fall a distance

x = 1/2 (-9.8 m/s²) (1 s)

x = 4.9 m

answer: i think he can see forward ten seconds and move forward ten seconds but giorno can undo his actions with ger.

explanation:

hi i’m sorry to call us today so we will get back with y’all and we will have to do it haha omg was the day