A 6 kg object is subjected to two forces, F1 = 17 N i - 9 N j and F2 = 16 N i -
ID: 1764667 • Letter: A
Question
A 6 kg object is subjected to two forces, F1 = 17 N i - 9 N j and F2 = 16 N i - 20 N j . The object is atrest at the origin at time t = 0. (a) What is the object's acceleration? ..........i m/s^2 +........j m/s^2(b) What is its velocity at time t = 14 s? ...........i m/s + .............j m/s
(c) Where is the object at time t = 14 s? ......... i m +........... j m A 6 kg object is subjected to two forces, F1 = 17 N i - 9 N j and F2 = 16 N i - 20 N j . The object is atrest at the origin at time t = 0. (a) What is the object's acceleration? ..........i m/s^2 +........j m/s^2
(b) What is its velocity at time t = 14 s? ...........i m/s + .............j m/s
(c) Where is the object at time t = 14 s? ......... i m +........... j m A 6 kg object is subjected to two forces, F1 = 17 N i - 9 N j and F2 = 16 N i - 20 N j . The object is atrest at the origin at time t = 0. (a) What is the object's acceleration? ..........i m/s^2 +........j m/s^2
(b) What is its velocity at time t = 14 s? ...........i m/s + .............j m/s
(c) Where is the object at time t = 14 s? ......... i m +........... j m
Explanation / Answer
Given : . Mass (m) = 6 kg . F1 = 17 N i - 9 N j and F2 = 16 N i - 20 N j . We know that : F = m a . ==> a = F /m . Here " F " is the resultant force : F = F1 - F2 ( when they areacting in opposite direction ) . and F = F1 + F2 ( when they are acting in the sane direction) . As the direction of forces is not mentioned here..... . I will consider F = F1 +F2 = ( 17 N i - 9 Nj ) + ( 16 N i - 20 N j) . = 33 i - 29 j . Hence acceleration is : F / m = (33 i - 29 j ) / 6 kg = 5.5 i + ( - 4.8 ) j m/s2 . (b) We know that : Acceleration (a) = V / t . ==> V = a * t = [ 5.5 i + ( - 4.8 ) j ] * 14 = 77 i + ( - 67.2 ) j m /s . (c) V = x / t . ==> x = V * t = [ 77 i + ( - 67.2 ) j ] * 14 = 1078 i + ( - 940.8 ) j m . Hope this helps u!Related Questions
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