A mass m 1 = 5 kg rests on a frictionless table and connected by a massless stri

Question

A mass m1 = 5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 3.6 kg. A force of magnitude F = 40 N pulls m1 to the left a distance d = 0.77 m.

1)How much work is done by the force F on the two block system? J

2)How much work is done by the normal force on m1 and m2? J

3)What is the final speed of the two blocks? m/s

4)How much work is done by the tension (in-between the blocks) on block m2? J

5)What is the tension in the string? N

6)The net work done by all the forces acting on m1 is:

positive

zero

negative

7)What is the NET work done on m1?

A mass m1 = 5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 3.6 kg. A force of magnitude F = 40 N pulls m1 to the left a distance d = 0.77 m. 1)How much work is done by the force F on the two block system? J 2)How much work is done by the normal force on m1 and m2? J 3)What is the final speed of the two blocks? m/s 4)How much work is done by the tension (in-between the blocks) on block m2? J 5)What is the tension in the string? N 6)The net work done by all the forces acting on m1 is: positive zero negative 7)What is the NET work done on m1?

Explanation / Answer

work done= F*d= 40*0.77== 30.8 j

work done by the normal force on m1 and m2= zero (since the normal force and distance moved are perpendicular to each other)

d = 0.77 m,u=0, a=F/m= 40/(5+3.6)== 4.65116279

so, using, v2 - u2 =2as

V2-0=2*4.651*0.77

V=final velocity=2.67629m/s

force on Mass m2= 3.6*4.651== 16.7436

tension in string=force on Mass m2== 16.7436 N

work done by the tension (in-between the blocks) on block m2=16.7436*0.77== 12.892572 J

net force acting on mass m1= 40-16.7436= 23.2564 N it is positive so net work done will be positive

Net work done = 23.2564*0.77== 17.907428 J

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