A mass m 4 kg. rests on a frictionless table and connected by a massless string

Question

A mass m 4 kg. rests on a frictionless table and connected by a massless string to another mass m2 4.8 kg. A force of magnitude F 26 N pulls m1 to the left a distance d 0.85 m 1) How much work is done by the force Fon the two block system? J Submit 22.1 2) How much work is done by the normal force on m1 and m2? J Submit 3) What is the final speed of the two blocks? 2.2 m/s Sul bmit 4) How much work is done by the tension (in-between the blocks) on block m2 J Submit 12 5) What is the tension in the string? N Submit 26 Your submissions 26 x Computed value: 26 Submitted Sunday, March 20 at 2:09 AM Feedback 6) The net work done by a the forces acting on m1 is: positive negative Submit 7) What is the NET work done on m12 J Submit

Explanation / Answer

m1= 4kg m2 = 4.8 kg F= 26 N

d= 0.85 m

a) work done by force = F*d = 26*0.85 = 22.1 J

b) 0 it is perpendiculer to velocity or motion

c) a = F /m = 26 / ( 4+4.8) = 2.95

v^2 = u^2 +2*a*s = 0 +2*2.95*0.85

v = 2.23 m/s

d)   

find the tension first

F - T = m1*a

T = F- m1*a = 26 - 4*2.95 = 14.2 N

work by tension on m1 = -T*d = -14.2* 0.85 = - 12.07

work done by tension on m2 = T*d = +12.07

total work = 0

E) T = F- m1*a = 26 - 4*2.95 = 14.2 N

F) positive

G) net work = F*d - T*d = (26-14.2) *0.85 = 10.03 J

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