Chapter 6, Problem 25P

To determine

(a) The tension in the cable

(b)The net workdone 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 if it started from rest.

Answer to Problem 25P

Solution:

(a) The tension in the cable is $3012.5\text{ N}$ .

(b)The net work done on the load is $7479.36\text{ J}$ .

(c) The work done by the cable on the load is $5422\text{5 J}$ .

(d) The work done by gravity on the load is $46746\text{ J}$ .

(e) The final speed of the load if it started from rest is $7.51\text{ m/s}$ .

Explanation of Solution

Free body diagram for given problem is

  

Given: Acceleration

  $\begin{array}{c}a=0.160g\\ =\left(0.160\right)\left(\text{9}{\text{.8 m/ s}}^{\text{2}}\right)\\ =1.568{\text{ m/s}}^{\text{2}}\end{array}$

Mass, $m=265\text{ kg}$

Distance, $d=18\text{m}$

a) To calculate the tension in the cable, we must use Newton's second law here. We have the force of gravity downward and force due to acceleration upward is

  $\begin{array}{l}F=ma\\ F=m\times (g+a)\\ F=265\times (9.8+1.568)\\ F=3012.5\text{ N}\end{array}$

The tension in the cable is 3012.5 N

b)

The net force and the distance that the load has traveled because of that force to find the net work done.

  $\begin{array}{l}{F}_{net}=ma\\ F_{net}=265\times 1.568\text{N }\\ F_{net}=415.52\text{ N}\end{array}$

Now net work done is

  $\begin{array}{l}{W}_{net}=F_{net}\times d\\ W_{net}=415.52\times 18\text{J}\\ W_{net}=7479.36\text{ J}\end{array}$

The net work done on the load is 7479.36 J

c)To find the work done by the cable only, we need to use the force that the cable applied on the load. This is the force of tension on the cable.

  $\begin{array}{l}W=F\times d\\ W=3012.5\times 18\\ W=54,225\text{ J}\end{array}$

To find the work done by the cable is 54225 J

d)The work done by gravity is

  $\begin{array}{l}W=-F_g\times d\\ W=-m\times g\times d\\ W=-265\times 9.8\times 18\\ W=-46746\text{ J}\end{array}$

This is a negative quantity because the displacement is up, but the force of gravity is down. 

e)Finding the final speed of the load

We know that the initial velocity is zero and use the Work-Energy Principle

  $W_{net} =\text{K}{\text{.E}}_f -\text{ K}{\text{.E}}_i$

  $W_{net} =\frac{1}{2}m{v}_f{}^2 -\frac{1}{2}m{v}_i{}^2$

  $W_{net} =\frac{1}{2}m{v}_f{}^2$

  $\begin{array}{l}{v}_f{}^2=\frac{2W_{net}}{m}\\ v_f=\sqrt{\frac{2W_{net}}{m}}\\ v_f=\sqrt{\frac{2\times 7479.36}{265}}\\ v_f=7.51\text{ m/s}\end{array}$

Final Speed of the Load is 7.51 m/s.

Conclusion:(a) The tension in the cable is 3012.5 N.

(b)The net work done on the load is 7479.36 J.

(c) The work done by the cable on the load is 54225 J.

(d) The work done by gravity on the load is 46746 J.

(e) The final speed of the load assuming that it started from rest is 7.51 m/s.