A mass m 1 = 4.9 kg rests on a frictionless table and connected by a massless st

Question

A mass m1 = 4.9 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.6 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.79 m.

1)How much work is done by gravity on the two block system?

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

3)What is the final speed of the two blocks?

4)How much work is done by tension on m1?

5)What is the tension in the string as the block falls?

6)The work done by tension on only m2is:

A.)positive

B.)zero

C.)negative

7) What is the NET work done on m2?

PLEASE show work step by step

Explanation / Answer

let,

mass, m1=4.9 kg

mass, m2=4.6 kg

distance, d=0.79 m

1)

work done by the gravity, w=m2*g*d

=4.6*9.8*0.79

=35.61 J

2)

work done by normal force is zero (0)

3)

use enegry relation,

1/2*(m1+m2)*v^2=m2*g*d

1/2*(4.9+4.6)*v^2=4.6*9.8*0.79

===> final speed, v=2.74 m/sec

4)

now,

m2*a=m2*g-T

and

m1*a=T

then,

===>

m2*a=m2*g-m1*a

===> a=m2*g/(m1+m2)

a=4.6*9.8/(4.9+4.6)

a=4.74 m/sec^2

acceleration, a=4.74 m/sec^2

now, tension T=m1*a

T=4.9*(4.74)

T=23.23 N

6)

work done by the tension on m2 is,

W=T*d

=23.23*0.79

=18.35 J

its is a positive work

7)

net work done, W_net=1/2*m2*v^2

=1/2*4.6*2.74^2

=17.27 J

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