Question

Two blocks of equal mass are connected by a string that passes over a pulley, as shown in the figure. Block 1 hangs from the string, and block 2 can slide on a table. The system is released from rest. Using conservation of energy, a student derives an expression for the speed v of block 1 when it has fallen a distance h after the system has been released from rest and obtains the equation v=gh−−√ . Question The figure presents a diagram of a pulley system at the edge of a table. A block labeled Block 1 is hanging off the table. Two other blocks are on the table. The block on the left is labeled Block 3 and the block on the right is labeled block 2. Block 1 is connected to a string that passes through a pulley. The left end of the string is connected to the right side of Block 2. Another string connected the left side of Block 2 to the right side of Block 3. A third block of the same mass as blocks 1 and 2 is attached to block 2 on the table, as shown in the figure. Using conservation of energy, the student repeats a derivation for the speed v of block 1 when it has fallen a distance h . Which of the following is a correct expression for v ? Responses gh√3 the fraction with numerator the square root of g h, and denominator 3 gh√2

Answer 1

The expression for the speed v of block 1 when it has fallen a distance h after the system has been released from rest is v = √(2gh), derived using conservation of energy.

Step-by-step explanation:

The expression for the speed v of block 1 when it has fallen a distance h after the system has been released from rest can be derived using conservation of energy.

1. Initially, block 1 has gravitational potential energy mgh and no kinetic energy.
2. When block 1 falls a distance h, it loses potential energy mgh and gains kinetic energy 1/2mv2.

Applying the principle of conservation of energy, we have:

mgh = 1/2mv2

Simplifying the equation, we find:

v = √(2gh)