Chapter 6, Problem 12P

A grocery cart with mass of 16 kg is being pushed at constant speed up a 12^o ramp by a force F_pwhich acts at an angle of 17^o below the horizontal. Find the work done by each of the forces $\left(m{\stackrel{\to }{g}}_{,}{\stackrel{\to }{F}}_N,{\stackrel{\to }{F}}_p\right)$ on the cart if the ramp is 7.5 m long.

To determine

The work done by the gravitational forces on the cart

Answer to Problem 12P

Solution:The work done by the gravitational forces on the cart is -244.5 J

Explanation of Solution

The various forces acting on the cart are shown in the figure.

  

Given: Mass of the cart, $m=16\text{ kg}$

Angle of inclination of ramp, $\theta =12°$

Length of ramp, $l=7.5\text{ m}$

Formula used:

  $\text{Work}\text{done}=\text{Force}\times \text{displacement}$

Calculation:

Work done the gravitational forces on the cart is

  $W_{mg}=mgd\sin 12°$

  $\begin{array}{c}{W}_{mg}=16\times 9.8\times 7.5\times \sin 12°\\ =244.5\text{}\text{J}\end{array}$

The negative sign indicates that weight opposes the upward motion.

Conclusion:The work done by the gravitational forces on the cart is $-244.5\text{ J}\text{.}$

To determine

The work done by the normal forces on the cart.

Answer to Problem 12P

Solution:The work done by the normal forces on the cart is zero

Explanation of Solution

The various forces acting on the cart are shown in the figure.

  

Given: Mass of the cart, $m=16\text{ kg}$

Angle of inclination of ramp, $\theta =12°$

Length of ramp, $l=7.5\text{ m}$

Formula used:

  $\text{Work}\text{done}=\text{Force}\times \text{displacement}$

Calculation:

The normal force is acting perpendicular to the ramp, θ = 90^o.

  $W_N=F_{N}d\cos 90=0$

The work done by the normal forces on the cart is zero

Conclusion:The work done by the normal forces on the cart is zero

To determine

The work done by the applied forces on the cart

Answer to Problem 12P

Solution:The work done by the applied forces on the cart is 244.5 J

Explanation of Solution

The various forces acting on the cart are shown in the figure.

  

Given: Mass of the cart, $m=16\text{ kg}$

Angle of inclination of ramp, $\theta =12°$

Length of ramp, $l=7.5\text{ m}$

Formula used:

  $\text{Work}\text{done}=\text{Force}\times \text{displacement}$

Calculation:

The work done by the applied force on the cart is

  $W_P=F_{P}d\cos 29$

  $\begin{array}{c}{W}_P=\left(mg\sin 12°\right)\left(d\right)\\ =16\times 9.8\times \sin 12°\times 7.5\\ =244.5\text{J}\end{array}$

Conclusion:The work done by the applied forces on the cart is 244.5J.