As shown in the figure below, two masses m 1 = 5.90 kg and m 2 which has a mass

Question

As shown in the figure below, two masses m1 = 5.90 kg and m2 which has a mass 80.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If m1 and m2 start from rest, after they have each traveled a distance h = 2.90 m, use energy content to determine the following.

(a) speed v of the masses

Question Part Points Submissions Used Question Part Points Submissions Used As shown in the figure below, two masses m1 = 5.90 kg and m2 which has a mass 80.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If m1 and m2 start from rest, after they have each traveled a distance h = 2.90 m, use energy content to determine the following. Question Part Points Submissions Us (a) speed v of the masses

Explanation / Answer

m1 = 5.90 kg

so, m2 = 80% of 5.90 kg

= 4.72 kg

Let the tension in the wire be T

So,

m1 * g - T = m1 * a    [ where a is the acceleration of block]

5.90 * 9.81 - T = 5.90 * a

T = 5.90 (9.81 - a) ------------(i)

and,

T - m2 * g = m2 * a

T = m2 ( g + a)

T = 4.72 ( 9.81 + a) ----------(ii)

Equating eq,i and eq. ii

5.90 (9.81 - a) = 4.72 (9.81 + a)

(5.90 + 4.72) * a = (5.90 - 4.72) * 9.81

a = 1.09 m/s2

Now,

V2 - U2 = 2as

As U = 0

V2 = 2 * 1.09 * 2.90

V = 2.51 m/s

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