91Ó°ÊÓ

Mass is $750$kg

The girder increases its speed from $0.25$m/s to $0.75$m/s

Distance is $3.5$m

A force $F$that acted on an object with mass $m$is parallel to the direction of motion and causes displacement $∆s$for the object.

The kinetic energy of the system change due to this force and the change in the kinetic energy represents the work done $W$

localid="1648138183418" role="math" $W=\left(F_s∆s\cos \mathrm{θ}\right)J$ (1)

Where $∆s$is the displacement of the motion and $\mathrm{θ}$is the angle between the motion and the force

$F=mg$

Where $m$is the mass of the object and $g$is the gravitational acceleration. The force of gravity is exerted in the vertical axis

$W_{gravity}=mg∆s\cos \mathrm{θ}$ (2)

Now, put the values for $m,g,∆s$an $\mathrm{θ}=180°$in equation (2)

localid="1648138244020" role="math" $$\begin{gathered} W_{gravity}=mg∆s\cos \mathrm{θ} \\ =\left((750kg)(9.8m/s^2)(3.5m)\cos 180°\right)J \\ =-25.7×{10}^{3}J \\ =-25.7kJ \end{gathered}$$

The change in the kinetic energy $∆K$represent the work done $W$or we can call that the work done causes a change for the kinetic energy of the system

$W=K_f-K_i$ (3)

Now, we know that,

$K=\frac{1}{2}m{v}^2$

Now,

$$\begin{gathered} W_{tension}=K_f-K_i \\ =\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2 \\ =\frac{1}{2}m\left[{v_f}^2-{v_i}^2\right] \end{gathered}$$

Now, we put the values

localid="1648138281178" $$\begin{gathered} W_{tension}=\frac{1}{2}m\left[{v_f}^2-{v_i}^2\right] \\ =\left(\frac{1}{2}(750kg)\left[\left(0.75m/s\right)-(0.25m/s)\right]\right)J \\ =187.5J \end{gathered}$$

The cable exerts work done against the gravity, so the work done by the cable is

$$\begin{gathered} W_{tension}=187.5J-(-25.7×{10}^3) \\ =25.9×{10}^3 \\ =25.9J \end{gathered}$$