A mass m 1 = 5.5 kg rests on a frictionless table and connected by a massless st
ID: 1320870 • Letter: A
Question
A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 3 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.72 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 m2 is:
positive
zero
negative
7)What is the NET work done on m2?
A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 3 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.72 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 m2 is: positive zero negative 7)What is the NET work done on m2?Explanation / Answer
Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.
A mass m1 = 3.8 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 5.5 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.77 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 m2 is:
positive
zero
negative
7) What is the NET work done on m2?
Answer
1) Work done by the gravity
W = m2 gH = 5.5*9.8*0.77 = 41.5 J
2)W = F.S = FS cosx
Work done is 0 as normal force is vertical and displacement is horizontal.
x is 90 deg.
3)Conservation of energy is to be applied.
m2 gH = (1/2)(m1+m2)v^2
v = 2.98 m/s
4)We need to find tension T
T = m1 a .....1
m2g - T = m2a .....2
T = 22.02 N
a = 5.8 m/s^2
Work done by T ==> T*0.77 = 16.96 J
5)T = 22.02 N
6) Negative
Angle b/w T and displaement of m2 is 180 deg.
W = FS cosx
7)Net work done on m2 = (m2g - T)*S
S is displacement.
W = 24.54 J
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