Tag Archives: weight

College Physics by Openstax Chapter 7 Problem 3

(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the elevator car by the gravitational force in this process? (c) What is the total work done on the elevator car?

Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=Fdcos⁡θW=Fd \cos \theta

Part A

The force in the cable is equal to the combined effect of the weight of the elevator and the friction that opposes the motion. That is

F=mg+fF=(1500 kg)(9.80 m/s2)+100 NF=14800 N\begin{align*} F & = mg + f \\ F & = \left( 1500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)+100\ \text{N} \\ F & = 14800\ \text{N} \end{align*}

This force in the cable is directed upward. The displacement is also upward, making the angle between the two quantities equal to zero. Thus, θ=0\theta = 0.

Substituting these values in the equation, the work done by the cable is

W=Fdcos⁡θW=(14 800 N)(40.0 m)cos⁡0∘W=592 000 JW=5.92×105 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 14\ 800\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 0^\circ \\ W & = 592\ 000\ \text{J} \\ W & = 5.92 \times 10^{5} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The force due to gravity is equal to the weight of the elevator alone. That is

Weight=mg=(1 500 kg)(9.80 m/s2)=14 700 N\begin{align*} \text{Weight} & = mg \\ & = \left( 1\ 500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \\ & = 14\ 700\ \text{N} \end{align*}

This force is directed downward, whereas the displacement is directed upward. Therefore, the angle θ\theta between the two quantities is θ=180∘\theta = 180^\circ.

Substituting these values in the formula for work, we have

W=Fdcos⁡θW=(14 700 N)(40.0 m)cos⁡180∘W=−588 000 JW=−5.88×105 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 14\ 700\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 180^\circ \\ W & = -588\ 000\ \text{J} \\ W & = -5.88 \times 10^{5}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Since the elevator is moving at a constant speed, it is in equilibrium. This means that the net external force experience by the elevator is zero. Therefore, the total work done on the elevator car is

WT=0 J  (Answer)W_{T} = 0\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

College Physics by Openstax Chapter 7 Problem 2

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)

Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is ΔPEg=mgh\Delta PE_{g} = mgh, with hh being the increase in height and gg the acceleration due to gravity.

W=mghW=mgh

We are given the following values: m=75.0 kgm=75.0\ \text{kg}, g=9.80 m/s2g=9.80\ \text{m/s}^2, and h=2.50 mh=2.50\ \text{m}.

Substitute the given in the formula.

W=mghW=(75.0 kg)(9.80 m/s2)(2.50 m)W=1837.5 NmW=1837.5 JW=1.84×103 J  (Answer)\begin{align*} W & = mgh \\ W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\ W & = 1837.5\ \text{Nm} \\ W & = 1837.5\ \text{J} \\ W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done is about 1.84×103 Joules1.84 \times 10 ^ {3}\ \text{Joules} .