Physics 8 fall 2015 Signed in as larry garcia Help Close Mo aw Question 5 Resour

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Physics 8 fall 2015 Signed in as larry garcia Help Close Mo aw Question 5 Resources previous | 5 of 5 | return to assignment Question 5 Part A Figure shows a block of mass m resting on a 20° slope The block has coefficients of friction -0.75 and k = 0.53 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg Figure 1) What is the minimum mass m that will stick and not slip? Express your answer to three significant figures and include the appropriate units m 1.91 kg Submit My Answers Give Up Correct If you need to use the rounded answer you submitted here in a subsequent part, instead use the full precision answer ision answer and only round as a final step before submitting an a nswer Part B If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have? Express your answer to three significant figures and include the appropriate units. a: 1 Value Units Figure 1 Submit My Answers Give Up Continue 20° 2 kg

Explanation / Answer

Part A)
Tension in the rope when the system of mass is at rest, is because of Hanging rope = 2 *9.8 N = 19.6 N

Now for mass m not to slip , Tension should be equal to -
mg*sin(20) + us*mg*cos(20) = T
m*9.8 * sin(20) + 0.75 * m*9.8*cos(20) = 19.6

Solving the above equation for m ,
m = 1.91 kg

Part B)

Let the acceleration be a.
For the Hanging mass we can write,
M*g - T = M*a
2*9.8 - T = 2 *a
T = 19.6 - 2a ---------1

Equation for the mass on incline ,

T - mg*sin(20) - uk*mg*cos(20) = m*a
T - 1.91*9.8*sin(20) - 0.53*1.91*9.8*cos(20) = 1.91*a --------2

Substituing Value of T from 1 into 2

19.6 - 2a - 1.91*9.8*sin(20) - 0.53*1.91*9.8*cos(20) = 1.91*a

Soving the above equation for calue of acceleration a, We get
a = 0.99 m/s^2

Acceleration, a = 1 m/s^2

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