Answer to Question #107145 in Classical Mechanics for Samson

Question

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%)

Answer 1

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