Find the acceleration reached by each of the two objects shown if the plane has a friction coefficient equal to 0.2. **Problem Statement:**Find the acceleration reached by each of the two objects shown if the plane has a friction coefficient of 0.2.**Diagram Explanation:**The diagram illustrates two objects connected by a string passing over a pulley. One object (7.00 kg) is on an inclined plane with an angle of 37.0°, and the other object (12.0 kg) hangs vertically. The inclined plane has a friction coefficient of 0.2.

Name: Find the acceleration reached by each of the two objects shown if the
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: Find the acceleration reached by each of the two objects shown if the plane has a friction coefficient equal to 0.2. **Problem Statement:**Find the acceleration reached by each of the two objects shown if the plane has a friction coefficient of 0.2.*

Given Data Mass of the two objects are m1=7.00 kg, m2=12.0 kg Coefficient of friction is μk=0.2 Angle of inclination of the slope is θ=37.0° Let acceleration of the two objects is a and T is the tension in the cable. Diagram Free body diagram for the two masses are shown below. Calculation for acceleration Considering downward direction as negative and upward direction as positive we get the equations of motions as written below. Equation of motion of object 1 along the vertical is, N-m1gcosθ=0 ⇒N=m1gcosθ So frictional force on the on object 1 is fk=μkN=μkm1gcosθ Equation of motion of the object 1 along the horizontal is, T-m1gsinθ-fk=m1a ⇒T-m1gsinθ-μkm1gcosθ=m1a .....................(1) Equation of motion of the object 2 along the vertical is, m2g-T=m2a .............................(2) Adding equation 1 and equation 2 we get, m1a+m2a=T-m1gsinθ-μkm1gcosθ+m2g-T ⇒(m1a+m2)×a=m2g-m1gsinθ-μkm1gcosθ ⇒a=m2g-m1gsinθ-μkm1gcosθm1+m2 ⇒a=m2g-m1g(sinθ+μkcosθ)m1+m2 substituting the values of m1, m2, μkand g we get, ⇒a=(12.0×9.80)-7.00×9.80×(sin 37.0°+0.2 cos 37.0°)(7.00+12.0) m/s2 ⇒a=117.6-(68.6×0.76)19.0 m/s2 ⇒a=117.6-52.119 m/s2 ⇒a=65.519 m/s2 ⇒a=3.45 m/s2 Answer Acceleration of the two objects is 3.45 m/s2