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Problem 2 (25 points) Two blocks (of masses ( m_{A} ) and ( m_{B} ) ) are connected by a massless string that passes over a pulley, as shown below. The coefficient of kinetic friction between ( m_{lambda} ) and the inclined plane (at angle ( heta ) above the horizontal) is ( mu_{k} ). When the system is released from rest, ( m_{B} ) moves upward and ( m_{A} ) moves down the incline. (a) Indicate clearly the direction of all forces acting on each block. You may use the figure above. ( 6 points) (b) If ( m_{A}=8.0 mathrm{~kg}, m_{B}=3.0 mathrm{~kg}, heta=50^{circ} ), and ( mu_{k}=0.3 ), determine the (common) acceleration of the two blocks and the tension in the string. (To get full credit, you must obtain this result from free-body diagrams and Newton's Second Law; simply plugging numbers into a formula derived in class is not enough! Set up appropriate force equations for each block, and solve these equations.) (19 points)

          Problem 2 (25 points)
Two blocks (of masses ( m_{A} ) and ( m_{B} ) ) are connected by a massless string that passes over a pulley, as shown below. The coefficient of kinetic friction between ( m_{lambda} ) and the inclined plane (at angle ( 	heta ) above the horizontal) is ( mu_{k} ). When the system is released from rest, ( m_{B} ) moves upward and ( m_{A} ) moves down the incline.
(a) Indicate clearly the direction of all forces acting on each block. You may use the figure above. ( 6 points)
(b) If ( m_{A}=8.0 mathrm{~kg}, m_{B}=3.0 mathrm{~kg}, 	heta=50^{circ} ), and ( mu_{k}=0.3 ), determine the (common) acceleration of the two blocks and the tension in the string. (To get full credit, you must obtain this result from free-body diagrams and Newton's Second Law; simply plugging numbers into a formula derived in class is not enough! Set up appropriate force equations for each block, and solve these equations.) (19 points)

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Problem 2 (25 points)
Two blocks (of masses ( mA ) and ( mB ) ) are connected by a massless string that passes over a pulley, as shown below. The coefficient of kinetic friction between ( mlambda ) and the inclined plane (at angle ( heta ) above the horizontal) is ( muk ). When the system is released from rest, ( mB ) moves upward and ( mA ) moves down the incline.
(a) Indicate clearly the direction of all forces acting on each block. You may use the figure above. ( 6 points)
(b) If ( mA=8.0 mathrm kg, mB=3.0 mathrm kg, heta=50^circ ), and ( muk=0.3 ), determine the (common) acceleration of the two blocks and the tension in the string. (To get full credit, you must obtain this result from free-body diagrams and Newton's Second Law; simply plugging numbers into a formula derived in class is not enough! Set up appropriate force equations for each block, and solve these equations.) (19 points)

Added by Stephen C.

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl… 10th Edition

Chapter 6

Instant Answer

Solved by Expert Vishal Gupta

10/16/2024

Step 1

- For block \( m_A \) on the incline: - Gravitational force: \( m_A g \) (acts downward). - Normal force: \( N \) (perpendicular to the incline). - Frictional force: \( f_k = \mu_k N \) (opposes motion, acts up the incline). - Tension: \( T \) (acts up Show more…

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Two blocks (of masses m_A and m_B) are connected by a massless string that passes over a pulley, as shown below. The coefficient of kinetic friction between m_A and the inclined plane (at angle θ above the horizontal) is μ_k. When the system is released from rest, m_B moves upward and m_A moves down the incline.

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00:01 Hi, here in this given problem, this is a rough inclined plane, having an angle of inclination theta, a box, box a kept over it, having mass m .a, a string passing over the pulley, holding another block, block b, having mass mb.

00:33 Its weight acting vertically down, mbg, tension t in the string, t will remain same.

00:44 Weight of the block a, m, a, g, that is also acting vertically down, two components of this weight.

00:53 This angle is theta.

00:54 So, component of weight m -a, that is m -a -g, that is m -ag, cost theta, perpendicular to the inclined plane, and m -a -g -g -sin -theta along the inclined plane in downward direction.

01:09 Normal reaction exerted by the inclined plane over the block a.

01:14 As the block b is having a tendency to move up, suppose it is moving up with an acceleration so block a will be sliding down with the same acceleration.

01:25 As it is sliding down, so force of friction will be acting on it up along the incline.

01:31 Fk.

01:32 So in the first part of the problem we have to show the direction of all the forces.

01:36 So direction of all the forces that is shown in the figure alongside.

01:57 Now in the second part of the problem, we have to find acceleration, a and tension t in the stream.

02:05 For which mass of block a, ma, 8 .0 kilogram, mass of block b means mb, this is 3 .0 kilogram, angle theta, 50 degree, coefficient of kinetic friction mu, mu, k, 0 .3, normal reaction using newton's third law of motion, it will be equal to mag, cost theta and fk, force of friction, kinetic friction, it will be equal to mu k times.

02:37 Normal reaction means mu k times mag, cost theta...

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