The acceleration experienced by each of the two objects, if the coefficient of kinetic friction between the 7kg object and the plane is 0.25 is 5.39m/s²
To solve this problem, we will use Newton's second law of motion. According to the law:
F = ma where:
F is the applied force
m is the mass
a is the acceleration of the body
For the body on the inclined, the sum of force on the body m1 is expressed as:
T - μR = m1a ............ 1
T - μmgcosθ = m1a ............ 1
For the body hanging, the sum of force on the body m2 is expressed as:
m2g - T = m2a ..............2
T is the tension
μ is the coefficient of friction
m1 and m2 are the masses
θ is the angle of inclination
a is the acceleration
Adding both equation will give:
- μm1gcosθ + m2g = (m1+m2)a
substitute the given values into the formula
-0.25(7)(9.8)cos 28 + 12(9.8) = (7+12)a
-17.15cos 28 + 117.6 = 19a
-15.143+117.6 = 19a
102.457 = 19a
a = 102.457/19
a = 5.39m/s²
Hence the acceleration experienced by each of the two objects, if the coefficient of kinetic friction between the 7kg object and the plane is 0.25 is 5.39m/s².
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