How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.33? Assume no friction acts on the wagon.
Solution:
The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle θ between them. In symbols,
W=Fdcosθ
In this case, we are given the following values:
Fdθ=50N=30m=30∘
Substituting these values into the equation, we have
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 20.0º with the horizontal. (See Figure 7.32.) 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.
Solution:
The Work Done by the Man on the Crate
The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle θ between them. In symbols,
W=Fdcosθ
In case where the work done by the man to the crate, the following values are given:
F=d=θ=500N4m0∘(Force is parallel to displacement)
Substituting these values in the equation, we have
W=W=W=Fdcosθ(500N)(4m)cos0∘2000N⋅m
The work done by the man on his body
In this case, the force exerted is counteracted by the weight of the man. This force is directed upward. The displacement is still the 4.0 m along the inclined plane. The angle between the force and the displacement is 70 degrees.
Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?
Solution:
Part A
According to Table 7.1, the energy in 1 gallon of gasoline is 1.2×108J. Since only 30% of the gasoline goes into useful work, the work done by the friction Wf is
WfWf=0.30(2.0gal)(1.2×108J/gal)=72×106J
Now, the work done by the friction can also be calculated using the formula below, where Ff is the magnitude of the friction force that keeps the car moving at constant speed, and d is the distance traveled by the car.
Wf=Ffd
We can solve for Ff in terms of the other variables.
Ff=dWf
Substituting the given values, we can now solve for the unknown magnitude of the force exerted to keep the car moving at constant speed.
If the required force is directly proportional to speed, then there must be a linear relationship between the required force and speed. In this situation, we can just simply used ratio and proportion to compute for the number of gallons.
(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 W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle θ between them. In symbols,
W=Fdcosθ
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
FFF=mg+f=(1500kg)(9.80m/s2)+100N=14800N
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.
Substituting these values in the equation, the work done by the cable is
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
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, with h being the increase in height and g the acceleration due to gravity.
W=mgh
We are given the following values: m=75.0kg, g=9.80m/s2, and h=2.50m.
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.
Solution:
The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle θ between them. In symbols,
W=Fdcosθ
We are given the following values: F=5.00N, d=0.600m, and θ=0∘.
Substitute the given values in the formula for work.
The Speed of the Roller Coaster at the Top of the Loop
Problem:
Modern roller coasters have vertical loops like the one shown in Figure 6.35. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?
Solution:
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The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads
Problem:
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).
(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.
(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
Solution:
Part A
The formula for an ideally banked curve is
tanθ=rgv2
Solving for v in terms of all the other variables, we have
v=rgtanθ
For this problem, we are given
radius, r=100m
acceleration due to gravity, g=9.81m/s2
banking angle, θ=15.0∘
Substituting all these values in the formula, we have
The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts
Problem:
A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.
(a) At what angular velocity is the centripetal acceleration 10g if the rider is 15.0 m from the center of rotation?
(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle θ below the horizontal will the cage hang when the centripetal acceleration is 10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle 10g should be.)
Solution:
Part A
The centripetal acceleration, ac, is calculated using the formula ac=rω2. Solving for the angular velocity, ω, in terms of the other variables, we should come up with
Now, summing forces in the horizontal direction, taking into account that Fc is the centripetal force which is the net force. That is,
Fc=mac
We know that Fc is equal to the horizontal component of the force Farm. That is Fc=Farmcosθ. Therefore,
Farmcosθ=mac
Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at 10g. Also, we note that the weight w is equal to mg. So, we have
Farmcosθsinθwcosθsinθmgcosθ=mac=m(10g)=10mg
From here, we are going to use the trigonometric identity tanθ=cosθsinθ. We can also cancel m, and g since they can be found on both sides of the equation.
Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).
(a) Show that θ (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, θ=tan−1(v2/rg)
(b) Calculate θ for a 12.0 m/s turn of radius 30.0 m (as in a race).
Solution:
Part A
Let us redraw the given forces in a free-body diagram with their corresponding components.
The force N and Fc are the vertical and horizontal components of the force F.
If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for F, we have
∑FyFcosθ−mgFcosθF=0=0=mg=cosθmgEquation 1
If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have
∑FxFsinθFsinθ=mac=mac=mrv2Equation 2
We substitute Equation 1 to Equation 2.
Fsinθcosθmgsinθmgcosθsinθ=mrv2=mrv2=mrv2
We can cancel m from both sides, and we can apply the trigonometric identity tanθ=cosθsinθ. We should come up with
gtanθtanθθ=rv2=rgv2=tan−1(rgv2)(Answer)
Part B
We are given the following values:
linear velocity, v=12.0m/s
radius of curvature, r=30.0m
acceleration due to gravity, g=9.81m/s2
We substitute the given values to the formula of θ we solve in Part A.
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