Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 6, Problem 46SP

A 1200-kg car coasts from rest down a driveway that is inclined 20° to the horizontal and is 15 m long. How fast is the car going at the end of the driveway if (a) friction is negligible and (b) a friction force of 3000 N opposes the motion?

(a)

Expert Solution
Check Mark
To determine

The speed of the car, of 1200 kg mass, at the end of the driveway that is inclined 20° to the horizontal and is 15 m long, considering the frictional force is negligible.

Answer to Problem 46SP

Solution:

10 m/s

Explanation of Solution

Given data:

The mass of the car is 1200 kg.

The angle of incline of the driveway with the horizontal is 20°.

The length of the driveway is 15 m.

Formula used:

The expression for change in kinetic energy is written as

KE=12m(vf2vi2)

Here, m is the mass, vf is the final velocity, vi is the initial velocity, and KE is the change in kinetic energy.

The expression for change in potential energy is written as

PEG=mg(hfhi)

Here, g is the acceleration due to gravity, hf is the final height, hi is the initial height, and PEG is the change in potential energy.

The expression for work energy theorem is written as

Change in kinetic energy+change in potential energy=Work done

Explanation:

Draw the diagram for the motion of the car

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 6, Problem 46SP , additional homework tip  1

There is no frictional force acting on the car, so there is no work done on the car. Therefore,

Change in kinetic energy+change in potential energy=0KE+PEG=0

Substitute 12m(vf2vi2) for KE and mg(hfhi) for PEG

12m(vf2vi2)+mg(hfhi)=012vf2+ghf=12vi2+ghi

The car started from rest. The initial velocity is zero. Also, the final position of the car is on the horizontal. Therefore,

hf=0

Consider the diagram and the initial vertical distance covered is the sine component of the distance covered. Therefore,

hi=(15 m)sin20°=5.13 m

Substitute 5.13 m for hi, 0 m for hf, 0 m/s for vi, and 9.81 m/s2 for g

12vf2+(9.81 m/s2)(0 m)=12(0 m/s)2+(9.81 m/s2)(5.13 m)vf=(2)(50.3253)vf=10.03 m/svf10 m/s

Conclusion:

The speed of the car at the end of the driveway with negligible frictional force is 10 m/s.

(b)

Expert Solution
Check Mark
To determine

The speed of the car, of 1200 kg mass, at the end of the driveway that is inclined 20° to the horizontal and is 15 m long, considering the frictional force of 3000 N.

Answer to Problem 46SP

Solution:

5.1 m/s

Explanation of Solution

Given data:

The mass of the car is 1200 kg.

The angle of incline is 20°.

The distance to the horizontal is 15 m.

The frictional force is 3000 N.

Formula used:

The expression of change in kinetic energy is written as

KE=12m(vf2vi2)

Here, m is the mass, vf is the final velocity, vi is the initial velocity, and KE is the change in kinetic energy.

The expression of change in potential energy is written as

PEG=mg(hfhi)

Here, g is the acceleration due to gravity, hf is the final height, hi is the initial height, and PEG is the change in potential energy.

The expression of frictional work done is written as

W=Ffscosθ

Here, Ff is the frictional force, W is the work done, s is the displacement, and θ is the angle between the displacement and force.

The expression for the work energy theorem is written as

Change in kinetic energy+change in potential energy=Work done

Explanation:

Draw the diagram for the motion of the car

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 6, Problem 46SP , additional homework tip  2

Recall that the expression for the work energy theorem is written as

Change in kinetic energy+change in potential energy=Work doneKE+PEG=W

Substitute 12m(vf2vi2) for KE, Ffscosθ for W, and mg(hfhi) for PEG

12m(vf2vi2)+mg(hfhi)=Ffscosθ

The car started from rest. The initial velocity is zero. Also, the final position of the car is on the horizontal. Therefore,

hf=0

Consider the diagram and the initial vertical distance covered is the sine component of the distance covered. Therefore,

hi=(15 m)sin20°=5.13 m

The frictional force acts in the opposite direction of the displacement. Therefore, θ=180°.

Substitute 5.13 m for hi, 0 m for hf, 0 m/s for vi, 180° for θ, 3000 N for Ff, 1200 kg for m, 15 m for s, and 9.81 m/s2 for g

12(1200 kg)(vf2(0 m/s)2)+(1200 kg)(9.81 m/s2)(0 m5.13 m)=(3000 N)(15 m)cos180°600vf260390.36=45000vf=60390.3645000600vf5.1 m/s

Conclusion:

The speed of the car with a frictional force of 3000 N is 5.1 m/s.

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Chapter 6 Solutions

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

Ch. 6 - 34. A 1200-kg car going 30 m/s applies its brakes...Ch. 6 - 35. A proton (m = 1.67 × 10−27 kg) that has a...Ch. 6 - 36. A 200-kg cart is pushed slowly at a constant...Ch. 6 - 6.37 [II] Repeat Problem 6.36 if the distance...Ch. 6 - 38. A 50 000-kg freight car is pulled 800 m up...Ch. 6 - 39. A 60-kg woman walks up a flight of stairs that...Ch. 6 - 40. A pump lifts water from a lake to a large tank...Ch. 6 - 41. Just before striking the ground, a 2.00-kg...Ch. 6 - 42. A 0.50-kg ball falls past a window that is...Ch. 6 - 43. At sea level a nitrogen molecule in the air...Ch. 6 - 44. The coefficient of sliding friction between a...Ch. 6 - 6.45 [II] Consider the simple pendulum shown in...Ch. 6 - 46. A 1200-kg car coasts from rest down a driveway...Ch. 6 - 47. The driver of a 1200-kg car notices that the...Ch. 6 - 48. A 2000-kg elevator rises from rest in the...Ch. 6 - 49. Figure 6-8 shows a bead sliding on a wire. How...Ch. 6 - 50. In Fig. 6-8, h1 = 50.0 cm, h2 = 30.0 cm, and...Ch. 6 - 51. In Fig. 6-8, h1 = 200 cm, h2 = 150 cm, and at...Ch. 6 - 6.52 [I] Imagine a 60.0-kg skier standing still on...Ch. 6 - 53. Considering the skier in the previous problem,...Ch. 6 - 6.54 [II] Considering the skier in the previous...Ch. 6 - 55. A 10.0-kg block is launched up a 30.0°...Ch. 6 - 56. Calculate the average power required to raise...Ch. 6 - 57. Compute the power output of a machine that...Ch. 6 - 58. An engine expends 40.0 hp in propelling a car...Ch. 6 - 6.59 [II] A 1000-kg auto travels up a 3.0 percent...Ch. 6 - 60. A 900-kg car whose motor delivers a maximum...Ch. 6 - 6.61 [II] Water flows from a reservoir at the rate...Ch. 6 - 6.62 [II] Find the mass of the largest box that a...Ch. 6 - 6.63 [II] A 1300-kg car is to accelerate from rest...
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