Question

Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The object m1 is held atrest on the floor, and m2 rests on a fixed incline of ? = 41.0°. The objects are released from rest, and m2 slides 1.70 m down the slope of the incline in 3.40 s.

Answer 1

Given thatDistance d=1.70 mtime taken t=3.40 smasses m₁ = 4.00 kgm₂ = 9.00 kgAcceleration due to gravity g=9.80 m/s²Angle θ =41.0°Initial velocity is u= 0--------------------------------------------------------------------------From the kinematic relations,d = ut+(1/2)at²(1.70 m) = 0 +(1/2)a(3.40 s)²a = 0.294m/s²Let us consider T be the tension in the string.----------------------------------------------------------------For m₁ :From Newton's second lawΣF = m₁a
-m₁g+T = m₁a
T = m₁g+m₁a= m₁(g+a)
T = (4.00 kg)((9.8 m/s²)+(0.294 m/s²))
= 40.376N
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For m₂:Apply Newton's seond law,
m₂gsinθ - T - f = m₂af =m₂gsinθ -T - m₂a
But N = m₂gcosθTherefore, the coefficient of friction isμ = f/N= (m₂gsinθ - T-m₂a)/(m₂gcosθ)= ((9.00 kg )(9.80 m/s²)sin41.0°-(40.376N) -(9.00 kg )(0.294 m/s²))/((9.00 kg )(9.80 m/s²)cos41.0°)=0.22