Not graded practice questions Date: 12/15l2020 8:01:00 PMTwo blocks are attached by a string as shown in the figure. The blocks havemasses M, = 1.4 kg and M, = 2.5 kg. Block 1 resides at a distance r = 0.25 m from the center of thetable (the table is round and free to rotate without friction). The static friction coefficient between blockM, and the table is µ = 0.51. Block 2 hangs vertically below as shown in the diagram. Assume themassof the string that connects the two masses is negligible.M.Determine the speed at which block 1 will begin to slide outward along the surface of the turntable. Enter your answer in méters per second.

Answered: 8:01:00 PM Two blocks are attached by a…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter5: Newton's Laws Of Motion

Section: Chapter Questions

Problem 76PQ: Jamal and Dayo are lifting a large chest, weighing 207 lb, by using the two rope handles attached to...

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Not graded practice questions

Date: 12/15l2020 8:01:00 PM
Two blocks are attached by a string as shown in the figure. The blocks have
masses M, = 1.4 kg and M, = 2.5 kg. Block 1 resides at a distance r = 0.25 m from the center of the
table (the table is round and free to rotate without friction). The static friction coefficient between block
M, and the table is µ = 0.51. Block 2 hangs vertically below as shown in the diagram. Assume the
massof the string that connects the two masses is negligible.
M.
Determine the speed at which block 1 will begin to slide outward along the surface of the turntable. Enter your answer in méters per second.

Expert Solution

Step 1 :Introduction

In the system there are mainly three force acting the axis of motion , which are the tension force exerted by the mass $M_{2} = 2.5 k g$ , which is , $T = M_{2} g$ , the frictional force on mass $M_{1} = 1.4 k g$ , which is , $f_{s} = μ_{s} M_{1} g$ and the centrifugal force on mass $M_{1} = 1.4 k g$ , which is , $F_{c} = \frac{M_{1} v^{2}}{r}$ , where $r = 0.25 m$ is the radius , $g = 9.8 m / s^{2}$ is the acceleration due to gravity and $μ_{s} = 0.51$ is the coefficient of static friction . Thus if the mass $M_{1} = 1.4 k g$ is moving outward , the acceleration of the system will be the centrifugal acceleration , which is $a_{c} = \frac{v^{2}}{r}$ .

According to the Newton's second law ,,for the block to move at a constant speed , that acceleration will be zero, so the relation of the forces will be $F_{c} - (T + f_{s}) = 0$

Step 2 :Calculation of the velocity

Calculate the velocity of the block of mass $M_{1} = 1.4 k g$ by using the relation $F_{c} - (T + f_{s}) = 0$ , where $F_{c} = \frac{M_{1} v^{2}}{r}$ is the centrifugal force , $T = M_{2} g$ is the tension force , $f_{s} = μ_{s} M_{1} g$ is the frictional force , $M_{2} = 2.5 k g$ , $r = 0.25 m$ is the radius , $g = 9.8 m / s^{2}$ is the acceleration due to gravity and $μ_{s} = 0.51$ is the coefficient of static friction .

$\begin{array}{rcl} F_{c} - (T + f_{s}) & = & 0 \\ \frac{M_{1} v^{2}}{r} - (M_{2} g + μ_{s} M_{1} g) & = & 0 \\ \frac{M_{1} v^{2}}{r} & = & (M_{2} g + μ_{s} M_{1} g) \\ v^{2} & = & \frac{(M_{2} g + μ_{s} M_{1} g) r}{M_{1}} \\ v & = & \sqrt{\frac{(M_{2} g + μ_{s} M_{1} g) r}{M_{1}}} \\ = & \sqrt{\frac{((2.5 k g) \times (9.8 m / s^{2}) + (0.51) \times (1.4 k g) \times (9.8 m / s^{2})) \times (0.25 m)}{(1.4 k g)}} \\ \approx & 2.4 m / s \end{array}$

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