Asked by Mary
A large solid cylinder rolls without slipping down a 40 m long hill with a 30 degree slope. The cylinder then rolls a short distance over level ground and then off a 20 meter high sheer cliff. How far from the base of the cliff does the cylinder land? Ignore friction and air resistance
Answers
Answered by
Elena
The law of conservation of energy
PE =KE,
PE = m•g•h = m•g•s•sinα,
KE = KE1 + KE2 = m•v^2/2 + I•ω^2/2=
= m•v^2/2+ (m•R^2/2) •v^2/2•R^2 =
=3•m•v^2/4.
m•g•s•sinα =3•m•v^2/4.
v = sqrt(4• m•g•s•sinα/3) = 16.2 m/s.
H =gt^2/2, t = sqrt(2•H/g) = 2.02 s.
L =v•t = 16.2 • 2.92 = 32.7 m.
PE =KE,
PE = m•g•h = m•g•s•sinα,
KE = KE1 + KE2 = m•v^2/2 + I•ω^2/2=
= m•v^2/2+ (m•R^2/2) •v^2/2•R^2 =
=3•m•v^2/4.
m•g•s•sinα =3•m•v^2/4.
v = sqrt(4• m•g•s•sinα/3) = 16.2 m/s.
H =gt^2/2, t = sqrt(2•H/g) = 2.02 s.
L =v•t = 16.2 • 2.92 = 32.7 m.
Answered by
Mary
Thank you so much, Elena!
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