Q3: A block of mass m - 6.00 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Assume ha-6.20 m.) 3.20 m 2.00 m (a) Determine the block\'s speed at points B and C. (b) Determine the net work done by the gravitational force on the block as it moves from point point C. to
Solution
Speed at point B
ETot = KE + U
KE = (1/2) m v^2
U = m g h
ETot, A = 0 + (6 kg)(9.8 m/s^2)(6.2 m) = 364.56 J
ETot = 364.56 J = constant
ETot, B = (1/2)(6 kg)(vB^2) + (6 kg)(9.8 m/s^2)(3.2 m)
(1/2)(6 kg)(vB^2) + 188.16 J = 364.56 J
vB = 7.67 m/s
speed at point C
ETot, C = (1/2)(6 kg)(vC^2) + (6 kg)(9.8 m/s2)(2.0 m)
(1/2)(6 kg)(vC^2) + 117.6 J = 364.56 J
vC = 9.07 m/s
b)
WAC = m g h
WAC = (6 kg) ( 9.8 m/s2) (6.2 m - 2.0 m)
WAC = (6 kg) ( 9.8 m/s2) (4.2m)
WAC = 246.96 J
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Disha NEET Physics Guide for classes 11 and 12.pdf
Q3- A block of mass m - 6-00 kg is released from rest from point and s.docx
1. Q3: A block of mass m - 6.00 kg is released from rest from point and slides on the frictionless
track shown in the figure below. (Assume ha-6.20 m.) 3.20 m 2.00 m (a) Determine the block's
speed at points B and C. (b) Determine the net work done by the gravitational force on the block
as it moves from point point C. to
Solution
Speed at point B
ETot = KE + U
KE = (1/2) m v^2
U = m g h
ETot, A = 0 + (6 kg)(9.8 m/s^2)(6.2 m) = 364.56 J
ETot = 364.56 J = constant
ETot, B = (1/2)(6 kg)(vB^2) + (6 kg)(9.8 m/s^2)(3.2 m)
(1/2)(6 kg)(vB^2) + 188.16 J = 364.56 J
vB = 7.67 m/s
speed at point C
ETot, C = (1/2)(6 kg)(vC^2) + (6 kg)(9.8 m/s2)(2.0 m)
(1/2)(6 kg)(vC^2) + 117.6 J = 364.56 J
vC = 9.07 m/s
b)
WAC = m g h
2. WAC = (6 kg) ( 9.8 m/s2) (6.2 m - 2.0 m)
WAC = (6 kg) ( 9.8 m/s2) (4.2m)
WAC = 246.96 J
