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Two blocks, one 0.8 kg and the other 2.0 kg are connected by a massless string over a frictionless pulley. The coefficient of kinetic friction is 0.14, and the downward ramp angle is 60 degrees.

a. Determine the acceleration of the blocks.
b. Calculate the tension of the string.

Sagot :

abidemiokin

Answer:

a) 4.58m/s²

b) 10.454N

Explanation:

According to Newton second law, the following simultaneous equation will be applicable

T - Fm = m1a.... 1

W - T = m2a ....2

m1 and m2 are the masses

T is the tension of the string

W is the weight of mass m2

Fm is the force acting along the incline

Fm = m1gsin theta

W = m2g

a) Add both equation to eliminate tension T

W-Fm = m1a+m2a

m2g - m1gsin theta = (m1+m2)a

Substitute the given values

2(9.8)-0.8(9.8)sin60 = (2+0.8)a

19.6-6.79 = 2.8a

12.81 = 2.8a

a = 12.81/2.8

a = 4.58m/s²

Hence the acceleration of the blocks is 4.58m/s2

b) Get the Tension T

Substitute the acceleration gotten in (a) into equation 1 as shown:

From 1:

T - Fm = m1a

T = m1a+Fm

T = 0.8(4.58)+m1gsintheta

T = 0.8(4.58)+0.8(9.8)sin60

T = 3.664+6.79

T = 10.454N

Hence the tension in the string is 10.454N

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