Answer to Question #104127 in Classical Mechanics for Antonia Marie Macias

Question

A portable animal cage that weighs 9,856 N rests on the floor. How much work (in J) is required to lift it 7.9 m vertically, at constant speed?

Answer 1

Applying Newton's second Law to calculate the magnitude of the vertical force.

$F_{Net}=m\;a=F-W$

Where.

• The acceleration a
• The mass of the cage m
• The weight of the cage W=9,856
• Applied force F

Remember that the acceleration is zero because it moves at constant velocity.

$a=0\;\frac{m}{s^{2}}$

Thus.

$m\;a=F-W \\ m\;0=F-W \\ F=W$

The external force applied to raise the cage to constant velocity

$F=9,856\;N$

To calculate the work we use the following equation.

$W=F\;d\;cos\theta$

Where.

• The force applied is $F=9,856\;N$
• Displacement is $d=7.9\;m$
• The angle between displacement and force is $\theta=0^{0}$

Evaluating numerically.

$W=F\;d\;cos\theta \\W=9,856\;N\times 7.9\;m\times cos(0^{0}) \\W=77,862.4\;J$

The required work is $W=77,862.4\;J$