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%)

Expert's answer

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)

m1gsin⁡θ−T−μm1gcos⁡θ=0m_1g\sin\theta-T-\mu m_1 g\cos\theta=0

T=m1gsin⁡θ−μm1gcos⁡θ−−−−−−(ii)T=m_1g\sin \theta-\mu m_1 g \cos \theta------(ii)

From equation (i) and (ii)

m2g=m1gsin⁡θ−μm1gcos⁡θm_2 g=m_1g\sin \theta-\mu m_1 g \cos \theta

⇒m2=m1gsin⁡θ−μm1gcos⁡θg\Rightarrow m_2=\dfrac{m_1g\sin \theta-\mu m_1 g \cos \theta}{g}

⇒m2=m1sin⁡θ−μm1cos⁡θ\Rightarrow m_2=m_1\sin\theta-\mu m_1 \cos\theta

⇒m2=12sin⁡32−12×0.14cos⁡32\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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