Question

slader At a carnival, you can try to ring a bell by striking a target with a 10.8-kg hammer. In response, a 0.408-kg metal piece is sent upward toward the bell, which is 4.23 m above. Suppose that 18.8 percent of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings

Answer 1

The hammer must be moving at a speed of approximately 4.082 meters per second.

Step-by-step explanation:

According to the statement and based on Principle of Energy Conservation, change in gravitational potential energy experimented by the metal piece (
`U_(g,o)`), in joules, must be equal to 18.8 percent of the translational kinetic energy of the hammer (
`K_(h)`), in joules.

`U_(g, o) = 0.188\cdot K_(h)` (1)

By definitions of gravitational potential and translational kinetic energies, we expand (1):

`m_(o)\cdot g\cdot h = 0.188\cdot \left((1)/(2)\cdot m_(h)\cdot v^(2)\right)` (2)

Where:

`m_(o)` - Mass of the metal piece, in kilograms.

`g` - Gravitational acceleration, in meters per square second.

`h` - Distance travelled by the metal piece, in meters.

`m_(h)` - Mass of the hammer, in kilograms.

`v` - Initial speed of the hammer, in meters per second.

If we know that
`m_(o) = 0.408\,kg`,
`g = 9.807\,(m)/(s^(2))`,
`h = 4.23\,m` and
`m_(h) = 10.8\,kg`, then the initial speed of the hammer is:

`m_(o)\cdot g\cdot h = 0.188\cdot \left((1)/(2)\cdot m_(h)\cdot v^(2)\right)`

`10.638\cdot m_(o)\cdot g \cdot h = m_(h)\cdot v^(2)`

`v = 3.261\cdot \sqrt{\frac {m_(o)\cdot g \cdot h}{m_(h)}}`

`v = 3.261\cdot \sqrt{((0.408\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (4.23\,m))/(10.8\,kg) }`

`v \approx 4.082\,(m)/(s)`

The hammer must be moving at a speed of approximately 4.082 meters per second.