2. A block with mass m=1.90 kg is placed on an inclined plane with slope angle a=32.0. The inclined plan has kinetic friction coefficient 0.43. Block 1 is connected to a second hanging block with mass m2=5.50 kg by a light cord hanging over a small, frictionless pulley with negligible mass and no slipping (as shown in the figure). Assume that the mass m2 is falling and thus m, moves up the incline. Draw free-body diagrams for the two blocks, identifying all forces. Using Newton's laws applied to each body, solve for the acceleration of mi up the incline. m1

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Homework Question 2 with Diagram.
2. A block with mass m=1.90 kg is placed on an inclined plane with slope angle a=32.0°. The
inclined plan has kinetic friction coefficient 0.43. Block 1 is connected to a second hanging
block with mass m2=5.50 kg by a light cord hanging over a small, frictionless pulley with
negligible mass and no slipping (as shown in the figure). Assume that the mass mz is falling
and thus m, moves up the incline.
Draw free-body diagrams for the two blocks, identifying all forces. Using Newton's laws
applied to each body, solve for the acceleration of mi up the inclihe.
m.
Transcribed Image Text:2. A block with mass m=1.90 kg is placed on an inclined plane with slope angle a=32.0°. The inclined plan has kinetic friction coefficient 0.43. Block 1 is connected to a second hanging block with mass m2=5.50 kg by a light cord hanging over a small, frictionless pulley with negligible mass and no slipping (as shown in the figure). Assume that the mass mz is falling and thus m, moves up the incline. Draw free-body diagrams for the two blocks, identifying all forces. Using Newton's laws applied to each body, solve for the acceleration of mi up the inclihe. m.
Expert Solution
Step 1

Given,

two mass connected by string over an inclined surface 

Step 2

Free body diagram 

Advanced Physics homework question answer, step 2, image 1

Step 3

From diagram 

for m1

R=m1gcosαthen friction force f=μRf=μm1gcosαf=0.43*1.9*9.8*cos32.00f=6.80 Nand T=f+m1gsinα+m1aT=6.80+1.90*9.81*sin32.00+1.90aT=16.67+1.90a  ....(1)

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