Please note that this problem is almost identical to #72 of chapter 5 in this bo
ID: 1696180 • Letter: P
Question
Please note that this problem is almost identical to #72 of chapter 5 in this book. However, the question asked here is phrased differently than the book version. Also, the solution given on this site is either wrong, or our professor is wrong in the possible solutions he offers. I'm very confused so I hope someone has the patience to EXPLAIN how to do this problem and reach the solution rather than just rush through it. Thank you so much.
Sorry that my diagram is so awful but it works :
mass of A= 50kg
mass of B= 100kg
friction coefficient (of A) = 0.25
angle = 37 degrees
distance x that block is moved = 20m
Explanation / Answer
Given that mass of the block A is m_A = 50 kg
mass of the block B is m _B = 100 kg
coefficient of friction is u_k = 0.25
frictional force f = u_k * m_A g cos (37)
= 0.25 * 50 kg * 9.8 * cos ( 37 )
= 97.8 N
Work done by the frictional force is W = - f * d
= - ( 97.8 N ) * 20m
= 1956.6 J
Kinetic energy is W = m_A g *d sin (37 ) - m_B * g *d +1/2 m_A v ^2 +1/ 2 m_B v^2
1956.6 J = 50 * 9.8* 20 (sin 37 ) - 100 kg * 9.8* 20 m + 1/2 ( 50+100) v^2
v = 12.5 m/s
Total change in kinetic energy is E = 1/2 ( 50 kg ) (12.5 )^2 -0
= 3.91*10^3 J