Potential Energy and Kinetic Energy

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Manh
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Homework Statement


Two blocks are hung by a string draped over a pulley, a 1.5-kg block on the left and a 3.0-kg block on the right. The two blocks start out at rest and at the same height.

A. What is the change in the gravitational potential energy of the system of blocks and Earth when the 3.0-kg block has dropped 0.60 m ?
B. What is the change in the kinetic energy of the system between release and this instant?
C. What is the velocity of the 1.5-kg block at this instant?

Homework Equations


U = m*g*h
K = 1/2 m*v^2

The Attempt at a Solution


A. U = (3.0 kg)(9.8 m/s^2)(0.60)
= 17.64 J
I wonder if the answer should have a negative sign!

B. K = 17.64 J
Since U is converted to K, therefore K has the same value of U. Am I correct?

C. 17.64 = 1/2*(1.5)*v^2
v = 4.85 m/s
These are my works and answers. I hope somebody can help me check them. Thanks!
 
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Manh said:
A. U = (3.0 kg)(9.8 m/s^2)(0.60)
= 17.64 J
I wonder if the answer should have a negative sign!
Don't ignore the movement of the 1.5 kg mass. (As far as sign, when something lowers the change in gravitational PE is negative.)

Manh said:
B. K = 17.64 J
Since U is converted to K, therefore K has the same value of U. Am I correct?
Yes, but you have the wrong value. (See A.)

Manh said:
C. 17.64 = 1/2*(1.5)*v^2
v = 4.85 m/s
These are my works and answers. I hope somebody can help me check them. Thanks!
Realize that the total kinetic energy is for both masses.
 
Doc Al said:
Don't ignore the movement of the 1.5 kg mass. (As far as sign, when something lowers the change in gravitational PE is negative.)
A. U1 = (1.5 kg)(9.8 m/s^2)(0.6 m) = 8.82 J
U2 = (3 kg)(9.8 m/s^2)(-0.6 m) = -17.64 J
delta U = -17.64 J + 8.82 J = -8.82 J
Doc Al said:
Yes, but you have the wrong value. (See A.)
B. delta K = 8.82 J
Doc Al said:
Realize that the total kinetic energy is for both masses.
C. K = 1/2 m1*v1^2 + 1/2 m2*v2^2. Where K = 8.82 J. Is this correct?
 
wouldn't you end up with 2 variables v1 and v2?