Student: emilybendey2 @gmal.com My Accoun Chapter 7 ILW. Begin Date: 9/1 00:00 A

Question

Student: emilybendey2 @gmal.com My Accoun Chapter 7 ILW. Begin Date: 9/1 00:00 AM -Due Date: 1082 f End Date: 1 (20%) Problem 2: Consider a box of mass 0.5 kg being accelerated aloeg floor at 1.6 ms2 Randomized Variables x-1.6ms Ifthis block is being accelerated fee est fr at me period of 9 s, what us the net work, inyoules, being done on t? Grade Sanmary Potestial 100% cotan asinacos Atenmpts remaning 1 1 21 atan) acotan sino cosbO anho cotanho DegreesRadians Hits deduction per bit His maning Feedback: debeten per feedback

Explanation / Answer

given that initial velocity is Vo = 0 m/sec

acelaration is a = 1.6 m/s^2

time taken is t = 9 sec

mass m = 6.5 kg then

Work done is W = F*S

F = m*a = 6.5*1.6 = 10.4 N

S is the displacement

using kinematic equations

S = Vo*t +(0.5*a*t^2)

S = (0*t)+(0.5*1.6*9^2) = 64.8 m

so the answer is W = F*S = 10.4*64.8 = 674 J

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