\(\displaystyle KE_{total}=KE_{trans}+KE_{rot}\)

\(\displaystyle KE_{total}=.5mv^2+.5I\omega^2\)

\(\displaystyle I=m*r^2\)

\(\displaystyle \omega=\frac{v}{r}\)

Combining equations

\(\displaystyle KE_{total}=.5mv^2+.5*m*v^2\)

Plugging in values:

\(\displaystyle KE_{total}=2.8*10^2\)

\(\displaystyle KE_{total}=280J\)

\(\displaystyle F=ma\)

Plugging in values

\(\displaystyle 1000=7500*a\)

\(\displaystyle a=.133\frac{m}{s}\)

Using

\(\displaystyle v_i+a*t=v_f\)

\(\displaystyle 0+.133*10=v_f\)

\(\displaystyle v_f=1.33\frac{m}{s}\)

Using

\(\displaystyle KE=.5mv^2\)

\(\displaystyle KE=.5*7500*1.33^2\)

\(\displaystyle KE=6633J\)

\(\displaystyle KE_{total}=KE_{trans}+KE_{rot}\)

\(\displaystyle KE_{total}=.5mv^2+.5I\omega^2\)

\(\displaystyle I=m*r^2\)

\(\displaystyle \omega=\frac{v}{r}\)

Combining equations

\(\displaystyle KE_{total}=.5mv^2+.5*m*v^2\)

Plugging in values:

\(\displaystyle KE_{total}=mv^2\)

\(\displaystyle KE_{total}=2.3*8^2\)

\(\displaystyle KE_{total}=147J\) 

Kinetic energy is equal to \(\displaystyle \frac{1}{2}mv^2\). The object of mass \(\displaystyle m\) and velocity \(\displaystyle v\) therefore has kinetic energy equal to \(\displaystyle \frac{1}{2}mv^2\). Let's let the object of mass \(\displaystyle 4m\) have velocity \(\displaystyle v_{4m}\). Therefore, its kinetic energy is \(\displaystyle \frac{1}{2}mv_{4m}^2\). We want to find \(\displaystyle v_{4m}\) such that the two objects to have the same kinetic energy, so we can equate their two kinetic energies. \(\displaystyle \frac{1}{2}mv^2=\frac{1}{2}4mv_{4m}^2\)

\(\displaystyle v^2=4v_{4m}^2\)

\(\displaystyle \frac{v^2}{4}=v_{4m}^2\)

\(\displaystyle \frac{v^2}{2}=v_{4m}\)

The equation for kinetic energy is:

\(\displaystyle KE=\frac{1}{2}mv^2\), where \(\displaystyle v\) is the velocity of the object. Therefore, if we halve the velocity, we can substitute that into our equation, and see what will change in the kinetic energy equation:

\(\displaystyle KE=\frac{1}{2}m(\frac{1v}{2})^2=\frac{1}{2}m\frac{1}{4}v^2\)

Thus, the kinetic energy is quartered.

To answer this question, we'll need to use the formula for kinetic energy.

\(\displaystyle KE=\frac{1}{2}mv^{2}\)

Plugging in the values given to us, we can solve for our answer.

\(\displaystyle KE=\frac{1}{2}(2.5\: kg)(4\: \frac{m}{s})^{2}\)

\(\displaystyle KE=20\:J\)

The equation for kinetic energy is:

\(\displaystyle KE=\frac{1}{2}mv^2\)

Where \(\displaystyle KE\) is the object's kinetic energy, \(\displaystyle m\) is the object's, and \(\displaystyle v\) is the object's velocity.

We can see that the relationship between kinetic energy and mass is linear, so when the velocity is doubled, the kinetic energy is also doubled.

The equation for kinetic energy is:

\(\displaystyle KE=\frac{1}{2}mv^2\)

Where \(\displaystyle KE\) is the object's kinetic energy, \(\displaystyle m\) is the object's, and \(\displaystyle v\) is the object's velocity.

We can see that the relationship between kinetic energy and velocity is quadratic, so when the velocity is doubled, the kinetic energy is quadrupled.

Find the final velocity by multiplying the average velocity by 2. Then substitute into the equation for kinetic energy.

\(\displaystyle V_{f}=\frac{110}{6}\times2=36.67\frac{m}{s}\)

\(\displaystyle KE=.5(3kg)(36.67\frac{m}{s})^2\rightarrow \boldsymbol{2017J}\)

This problem can either be very simple or very complex.

You could convert the contractor's potential energy into kinetic energy, and then find the velocity at which he hits the seesaw. Then, use that to calculate the kinetic energy of the bricks and the height at which the bricks will fly. Although some may find that fun to do, it's unnecessary.

If we assume that all of the energy of the man is transferred to the bricks, we can use the conservation of energy equation:

\(\displaystyle E = U_i + K_i = U_f + K_f\)

We need to make a few clarifying statements for our initial and final states. In the initial state, the contractor is at a height of 1.5m and not moving. The bricks are not moving at this point either. In the final state, the man is on the ground not moving, and the bricks are at a height of 10m and not moving. Therefore, we can rewrite the above equation as the following:

\(\displaystyle m_cgh_i = m_bgh_f\)

Rearranging for the mass of the bricks:

\(\displaystyle m_b = m_c \frac{h_i}{h_f} = (80kg)(1.5m)(10m)= 12kg\)