Answer:
3.93 m/s
Explanation:
Let the kinetic energy of hammer be 'K' and speed of hitting the target be 'v'.
Given:
Mass of the hammer (M) = 7.76 kg
Mass of the metal piece (m) = 0.372 kg
Kinetic energy of the hammer used by the metal piece = 29.7% of 'K' = 0.297K
Vertical height traveled by the metal piece (h) = 4.87 m
From conservation of energy, the kinetic energy used by the metal piece is transformed to the gravitational potential energy when it reaches the height of the bell.
Gravitational potential energy of the piece is given as:
[tex]U=mgh\\\\U=0.372\times 9.8\times 4.87=17.754\ J[/tex]
Now, as per question:
[tex]0.297K=17.754\ J\\\\K=\frac{17.754}{0.297}\\\\K=59.78\ J[/tex]
Therefore, the kinetic energy of the hammer is 59.78 J.
We know that,
Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
So, [tex]K=\frac{1}{2}mv^2[/tex]
Expressing in terms of 'v', we get:
[tex]mv^2=2K\\\\v^2=\frac{2K}{m}\\\\v=\sqrt{\frac{2K}{m}}[/tex]
Plug in the given values and solve for 'v'. This gives,
[tex]v=\sqrt{\frac{2\times 59.78}{7.76}}\\\\v=3.93\ m/s[/tex]
Therefore, the hammer must move with a speed of 3.93 m/s when it strikes the target so that the bell just barely rings.