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(m1)=12kgA(m_1)=12 kg

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

Angle of inclination (θ)=32.0(\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 m2m_2

So,T=m2g(i)T=m_2 g-------(i)

m1gsinθTμm1gcosθ=0m_1g\sin\theta-T-\mu m_1 g\cos\theta=0

T=m1gsinθμm1gcosθ(ii)T=m_1g\sin \theta-\mu m_1 g \cos \theta------(ii)

From equation (i) and (ii)

m2g=m1gsinθμm1gcosθm_2 g=m_1g\sin \theta-\mu m_1 g \cos \theta

m2=m1gsinθμm1gcosθg\Rightarrow m_2=\dfrac{m_1g\sin \theta-\mu m_1 g \cos \theta}{g}

m2=m1sinθμm1cosθ\Rightarrow m_2=m_1\sin\theta-\mu m_1 \cos\theta

m2=12sin3212×0.14cos32\Rightarrow m_2=12\sin 32-12\times0.14\cos 32

m2=4.93kg\Rightarrow m_2=4.93kg

So, mass of block B will be 4.93 kg


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