(c) The position, in terms of time, of a block of mass 0.50 kg on a frictionless horizontal surface, at the end of a horizontal spring which has a force constant of 1000 N/m, is x = (0.022m)cos(2mft) (i) Determine the total mechanical energy of the system. (ii) Determine the maximum velocity of the block. (iii) Determine the velocity of the block when x=0.011m.

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter16: Oscillations
Section: Chapter Questions
Problem 64PQ: Use the data in Table P16.59 for a block of mass m = 0.250 kg and assume friction is negligible. a....
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(c) The position, in terms of time, of a block of mass 0.50 kg on a frictionless horizontal surface, at
the end of a horizontal spring which has a force constant of 1000 N/m, is
x = (0.022m)cos(2nft)
) Determine the total mechanical energy of the system.
(ii) Determine the maximum velocity of the block.
(iii) Determine the velocity of the block when x= 0.011 m.
(iv) What would the graph of x = (0.022m)cos[(2nft) + represent? [Discuss the similarities
and the differences between the two graphs of
x= (0.022m)cos(2nft) and
x = (0.022m)cos[(2ft) +.
(v) Draw the acceleration vs time graphs on top of cach other, of
x= (0.022m)cos(2nft) and
x = (0.022m)cos[(2mft) +.
Interpret what is represented graphically.
Transcribed Image Text:(c) The position, in terms of time, of a block of mass 0.50 kg on a frictionless horizontal surface, at the end of a horizontal spring which has a force constant of 1000 N/m, is x = (0.022m)cos(2nft) ) Determine the total mechanical energy of the system. (ii) Determine the maximum velocity of the block. (iii) Determine the velocity of the block when x= 0.011 m. (iv) What would the graph of x = (0.022m)cos[(2nft) + represent? [Discuss the similarities and the differences between the two graphs of x= (0.022m)cos(2nft) and x = (0.022m)cos[(2ft) +. (v) Draw the acceleration vs time graphs on top of cach other, of x= (0.022m)cos(2nft) and x = (0.022m)cos[(2mft) +. Interpret what is represented graphically.
Expert Solution
Step 1

(1) Total Mechanical Energy = 12kA2 = 12×1000×0.0222 = 0.242 J

Step 2

Maximum velocity will be at the mean position , at mean position the total mechanical energy will be in the form of kinetic energy.12mv2 =Total mechanical Energy12(0.5)v2 = 0.242 V = 0.9838 m/s .

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