Question

ph3u1 - unit 2 forces

1. assuming the coefficient of kinetic friction between ( m_1 ) and the table top is 0.38, and the object ( m_2 ) has a mass of 9.5 kg. if ( m_1 ) accelerates at a rate of ( 0.75 m/s^2 ) right, determine the mass ( m_1 ). 4 marks - t
2. jelene decides to share a drawing she made from the ( 3^{rd} ) grade. the locomotive is able to generate an applied force of 4800 n left. assume that there is no friction or air resistance. answer the following:

a. draw a fbd for the system, you may split it into individual objects. 2 marks - c
b. determine the acceleration of the train 3 marks - a
c. determine the tension force between cart 1 and cart 2. 2 marks - a

Explanation:

Step1: Analyze forces on \(m_2\)

According to Newton's second law \(F = ma\), for \(m_2\), \(m_2g - T=m_2a\). So \(T = m_2(g - a)\). Substitute \(m_2 = 9.5\space kg\), \(g=9.8\space m/s^2\), \(a = 0.75\space m/s^2\).
\(T=9.5\times(9.8 - 0.75)=9.5\times9.1 = 86.45\space N\)

Step2: Analyze forces on \(m_1\)

For \(m_1\), \(T-\mu_k m_1g=m_1a\). Rearrange the formula to solve for \(m_1\): \(m_1=\frac{T}{\mu_k g + a}\)
Substitute \(T = 86.45\space N\), \(\mu_k=0.38\), \(g = 9.8\space m/s^2\), \(a=0.75\space m/s^2\)
\(m_1=\frac{86.45}{0.38\times9.8+0.75}=\frac{86.45}{3.724 + 0.75}=\frac{86.45}{4.474}\approx19.3\space kg\)

Answer:

\(m_1\approx19.3\space kg\)