atwoods machineinclinefrictionKinematic Equationspulley

(Mid Term 1) Two blocks are connected by a string and pulley as shown. Assuming that the string and the pulley are massless, the magnitude of the acceleration of each block is?
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Solution

According to Newton's Second Law

∑Fm1=T−m1g=m1a∑Fm2=m2g−T=m2a⇒a=m2−m1m1+m2⋅g=110−9090+110⋅9.8=0.98m/s2\begin{aligned} \sum F_{m1} &= T-m_1 g = m_1a\\ \sum F_{m2} &= m_2 g - T = m_2a\\ \To & a = \frac{m_2-m_1}{m_1+m_2} \cdot g \\ &= \frac{110-90}{90+110} \cdot 9.8 = 0.98m/s^2 \end{aligned}

m1=100.0kg,m2=40.0kgm_1 = 100.0 kg,m_2=40.0kg and θ=30°\th = 30\degree. The coefficient of kinetic friction between the block of mass m1m_1 and the incline is μk=0.70\mu_k = 0.70. The initial velocity of m1m_1 is 1.5m/s1.5m/s. At what distance will m1m_1 stop?
(m1=100kgm_1=100kg)
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Solution

According to Newton's Second Law,

∑Fx=m1a=T−W1x−Fk∑Fy=0=N−W1yFk=N⋅μkW1x=W⋅sin⁡θW1y=W⋅cos⁡θ∑F2y=m2a=T−m2g\begin{aligned} \sum F_x &= m_1 a = T-W_{1x}-F_k\\ \sum F_y &= 0 = N - W_{1y} \\ F_k &= N \cdot \mu_k\\ W_{1x} &= W\cdot \sin \th\\ W_{1y} &= W\cdot \cos \th\\ \sum F_{2y} &= m_2 a = T - m_2 g \end{aligned}

Solve for aa, a=m2g−ukm1gcos⁡θ−m1gsin⁡θm1+m2≈−4.94m/s2a = \dfrac{m_2 g - u_k m_1 g \cos\th - m_1 g \sin \th}{m_1+m_2}\approx -4.94m/s^2.
According to Kinematic Equations,

vf2=v02+2ax⇒x=−v022a=−1.522×(−4.94)≈0.227m\begin{aligned} v_f^2 &= v_0^2 + 2ax \\ \To x &= -\frac{v_0^2}{2a}\\ &= -\frac{1.5^2}{2 \times (-4.94)} \approx 0.227m \end{aligned}