Answer to Question #107145 in Classical Mechanics for Samson

Question #107145
Show complete solution. Use the correct number of significant figures.

1. In the figure, two blocks are connected over a pulley. The mass of block A is 12.0 kg and the coefficient of kinetic friction between A and the incline is 0.140. Angle θ of the incline is 32.0°. Block A slides down the incline at constant speed. What is the mass of block B?

Answer: _______

(the tolerance is +/-2%)


2. In the figure, a 4.1 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is friction less until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.7 m/s, the height difference is h = 1.1 m, and μk = 0.583. Find d.

Answer:________

(the tolerance is +/-2%)
1
Expert's answer
2020-03-31T09:33:06-0400

As per the given question,

mass of the block "A(m_1)=12 kg"

Coefficient of kinetic friction "(\\mu)=0.14"

Angle of inclination "(\\theta)=32.0^\\circ"

Block A is slides with constant speed in downward direction,

now, let the tension in the string be T,and mass of the block B is "m_2"

So,"T=m_2 g-------(i)"

"m_1g\\sin\\theta-T-\\mu m_1 g\\cos\\theta=0"

"T=m_1g\\sin \\theta-\\mu m_1 g \\cos \\theta------(ii)"

From equation (i) and (ii)

"m_2 g=m_1g\\sin \\theta-\\mu m_1 g \\cos \\theta"

"\\Rightarrow m_2=\\dfrac{m_1g\\sin \\theta-\\mu m_1 g \\cos \\theta}{g}"

"\\Rightarrow m_2=m_1\\sin\\theta-\\mu m_1 \\cos\\theta"

"\\Rightarrow m_2=12\\sin 32-12\\times0.14\\cos 32"

"\\Rightarrow m_2=4.93kg"

So, mass of block B will be 4.93 kg


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