Question

Question 25: (III) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.160 g by a single cable. Determine (a) the tension in the cable; (b) the net work done on the load; (c) the work done by the cable on the load; (d) the work done by gravity on the load; (e) the final speed of the load assuming it started from rest.

Short answer

1. The tension in the cable is 3012.5 N.
2. Net work done on the load is 7479.4 J.
3. Work done by the cable on the load is 54,225 J.
4. Work done by the gravity on the load is 46,746 J.
5. The final speed of the load is \(7.51\;{\rm{m/s}}\).

Step-by-step solution

Work-energy principle

According to the work-energy principle, the net work done by an object is equal to the change in kinetic energy of that object, i.e.,

\(\begin{array}{c}\Delta KE = {W_{{\rm{net}}}}\\K{E_{\rm{f}}} - K{E_{\rm{i}}} = {W_{{\rm{net}}}}\end{array}\)

Given information

Mass of load is m = 265 kg.

Displacement of load in vertically upwards direction is s = 18.0 m.

Acceleration of the load is a = 0.160g.

Here, g is the acceleration due to gravity.

The initial speed of the load is \({v_{\rm{i}}} = 0\;{\rm{m }}{{\rm{s}}^{ - 1}}\).

Let the final speed of the load is \({v_{\rm{f}}}\).

Free-body diagram of the load

A free-body diagram of the load is shown in the figure below:

The forces acting on the load are as follows:

• Tension in the cable T acting vertically upwards.
• Force of gravity, \({F_{\rm{g}}} = m{\rm{g}}\) acting vertically downwards.
• Net force due to acceleration, \(F = ma\) acting vertically upwards.

(a) Determining tension in the cable

Let us take a vertically upward direction as the positive direction. From the figure, apply Newton’s second law in the vertical direction.

\(\begin{array}{c}T - m{\rm{g}} = ma\\T = ma + m{\rm{g}}\\ = m\left( {a + {\rm{g}}} \right)\\ = \left( {265\;{\rm{kg}}} \right) \times \left( {0.160{\rm{g}} + {\rm{g}}} \right)\\ = \left( {265\;{\rm{kg}}} \right) \times \left( {1.16 \times 9.8\;{\rm{m }}{{\rm{s}}^{ - 2}}} \right)\\ = 3012.5\;{\rm{N}}\end{array}\)

Thus, the tension in the cable is 3012.5 N.

(b) Determining net work done on the load

Since the direction of both the net force and the displacement of the load are vertically upward, the net work done on the load is as follows:

\(\begin{array}{c}{W_{{\rm{net}}}} = Fs\cos 0^\circ \\ = mas\cos 0^\circ \\ = \left( {265\;{\rm{kg}}} \right) \times \left( {0.16 \times 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {18.0\;{\rm{m}}} \right)\\ = 7479.4\;{\rm{J}}\end{array}\)

Thus, net work done on the load is 7479.4 J.

(c) Determining the work done by the cable on the load

Since the direction of the force acted by the cable (i.e., tension) and the displacement of the load are vertically upward, the work done by the cable on the load is as follows:

\(\begin{array}{c}{W_{{\rm{net}}}} = Ts\cos 0^\circ \\ = \left( {3012.5\;{\rm{N}}} \right) \times \left( {18.0\;{\rm{m}}} \right)\\ = 54,225\;{\rm{J}}\end{array}\)

Thus, work done by the cable on the load is 54,225 J.

(d) Determining the work done by gravity on the load

Since the direction of both the force of gravity and the displacement of the load are opposite to each other, the work done by gravity on the load is as follows:

\(\begin{array}{c}{W_{{\rm{net}}}} = {F_{\rm{g}}}s\cos 180^\circ \\ = m{\rm{g}}s( - 1)\\ = \left( {265\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {18.0\;{\rm{m}}} \right) \times \left( { - 1} \right)\\ = - 46,746\;{\rm{J}}\end{array}\)

Thus, work done by the gravity on the load is 46,746 J.

(e) Determining the work done by gravity on the load

According to the work-energy principle, change in kinetic energy of the load is equal to the net work done by the load, i.e.,

\(\begin{array}{c}K{E_{\rm{f}}} - K{E_{\rm{i}}} = {W_{{\rm{net}}}}\\\frac{1}{2}mv_{\rm{f}}^{\rm{2}} - \frac{1}{2}mv_{\rm{f}}^{\rm{2}} = {W_{{\rm{net}}}}\\\frac{1}{2}m(v_{\rm{f}}^{\rm{2}} - v_{\rm{i}}^{\rm{2}}) = {W_{{\rm{net}}}}\end{array}\)

On substituting values of variables in the above expression, you will get final velocity as follows:

\(\begin{array}{l}\frac{1}{2} \times \left( {265\;{\rm{kg}}} \right) \times (v_{\rm{f}}^{\rm{2}} - 0) = 7479.4\;{\rm{J}}\\v_{\rm{f}}^{\rm{2}} = 56.45\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\{v_{\rm{f}}} = 7.51\;{\rm{m/s}}\end{array}\)

Thus, the final speed of the load is \(7.51\;{\rm{m/s}}\).