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Chapter 7: Q.21CQ (page 259)

Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side. Is your increase in gravitational potential energy the same in both cases? Is your energy consumption the same in both?

Short Answer

Expert verified

A zigzag path is easier to climb a mountain than a straight path because it requires less energy. The gravitational potential energy is same in both paths. The energy consumptions per unit time is smaller within the zigzag path rather than one straight up the side.

Step by step solution

01

Definition of Concepts

Power: Power is a scaler quantity defined as the rate at which energy is consumed.

Gravitational potential energy: The gravitational potential energy stored in the body equals the work done in raising a body to some height against the gravitational force of attraction.

The gravitational potential energy is given as,

PE=mgh

Here, m is the mass of the body, g is the acceleration due to gravity and h is the height from which the surface object is raised.

The height of the mountain (displacement of the climber) is the same in both cases whether the climber takes a zigzag path or a straight path. Hence, the gravitational potential energy is the same in both cases.

02

Explain the energy consumption per unit of time is smaller in the zigzag rather than the straight path while climbing up a mountain

The power is given as,

Power=EnergyTime

The zigzag path is longer than the straight path because it takes longer to work while climbing a mountain.

From the expression of power, we can conclude that it takes less power to climb a mountain via a zigzag path rather than a straight path. This makes the climb easier on a zigzag path as it is easier to generate a small amount of power for long periods of time in a zigzag path rather than to generate a large power for a short period of time.

So, the energy consumption per unit of time is smaller in the zigzag rather than the straight path while climbing up a mountain.

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Most popular questions from this chapter

(a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months?

(b) How long can a battery that can supply 8.00脳104 J run a pocket calculator that consumes energy at the rate of 1.00脳10鈭�3 W?

Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-watt device.) Explain in terms the definition of power.

(a) Calculate the power per square meter reaching Earth鈥檚 upper atmosphere from the Sun. (Take the power output of the Sun to be\(4.00 \times {10^{26}}{\rm{ W}}\).)

(b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of\(1.30{\rm{ kW}}/{{\rm{m}}^2}\)reaches Earth鈥檚 surface. Calculate the area in\({\rm{k}}{{\rm{m}}^2}\)of solar energy collectors needed to replace an electric power plant that generates\(750{\rm{ MW}}\)if the collectors convert an average of\(2.00\% \)of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States鈥� energy needs\(\left( {1.05 \times {{10}^{20}}{\rm{ J}}} \right)\)? Australia鈥檚 energy needs\(\left( {5.4 \times {{10}^{18}}{\rm{ J}}} \right)\)? China鈥檚 energy needs\(\left( {6.3 \times {{10}^{19}}{\rm{ J}}} \right)\)? (These energy consumption values are from 2006.)

A car鈥檚 bumper is designed to withstand a \(4.0 - {\rm{km}}/{\rm{h}}\) \(\left( {1.1 - {\rm{m}}/{\rm{s}}} \right)\) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses \(0.200{\rm{ m}}\) while bringing a \(900 - {\rm{kg}}\) car to rest from an initial speed of \(1.1{\rm{ m}}/{\rm{s}}\).

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0潞 with the horizontal. (See Figure 7.35.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.35 A man pushes a crate up a ramp
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