Solution Found!
Two boxes, m1 = 1.0 kg with a coefficient of kinetic friction of 0.10, and m2 = 2.0 kg
Chapter 4, Problem 87GP(choose chapter or problem)
Two boxes, \(m_{1}=1.0 \mathrm{~kg}\) with a coefficient of kinetic friction of 0.10, and \(m_{2}=2.0 \mathrm{~kg}\) with a coefficient of 0.20, are placed on a plane inclined at \(\theta=30^{\circ}\).
(a) What acceleration does each box experience?
(b) If a taut string is connected to the boxes (Fig. 4-64 ), with \(m_{2}\) initially farther down the slope, what is the acceleration of each box?
(c) If the initial configuration is reversed with (m_{1}\) starting lower with a taut string, what is the acceleration of each box?
Questions & Answers
QUESTION:
Two boxes, \(m_{1}=1.0 \mathrm{~kg}\) with a coefficient of kinetic friction of 0.10, and \(m_{2}=2.0 \mathrm{~kg}\) with a coefficient of 0.20, are placed on a plane inclined at \(\theta=30^{\circ}\).
(a) What acceleration does each box experience?
(b) If a taut string is connected to the boxes (Fig. 4-64 ), with \(m_{2}\) initially farther down the slope, what is the acceleration of each box?
(c) If the initial configuration is reversed with (m_{1}\) starting lower with a taut string, what is the acceleration of each box?
Step 1 of 5
In this question, we need to find acceleration experienced by each box
In part b, we need to find the acceleration of each box if taut string is connected to boxes and \(m_{2}\) initially farther down the slope
In part c, we need to find the acceleration of each box if initial configuration is reversed with \(m_{1}\) starting lower with a taut string
Data given
Mass of box \(m_{1}=1.0 \mathrm{~kg}\)
Mass of box \(m_{2}=2.0 \mathrm{~kg}\)
Coefficient of kinetic friction for box \(m_{1}\) is \(\mu_{k 1}=0.10\)
Coefficient of kinetic friction for box \(m_{2}\) is \(\mu_{k 2}=0.20\)
Angle of inclination \(\theta=30^{0}\)