PHYS001- Work, Power and Energy PDF

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Work, Power and Energy...

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ASSIGNMENT #5: WORK, POWER AND ENERGY

1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the net work done on the block is 247 J. What angle does the rope make with the horizontal?

Given:

Δ x =7.0 m F=40 N W =247 J Aim:

Angle θ Solution:

W =F Δxcosθ

co s θ=

W FΔx

( FWΔ x )

θ=cos−1

7.0 m 247 J ( 40 N ) (¿¿) ¿ θ=cos−1 ¿ θ=28.1°

2. Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 45 N directed at an angle of 41° below the horizontal direction. Calculate the work that Mike does on the mower in pushing it 9.1 m across the yard. Given:

F=45 N

θ=41 ° below the horizontal direction Δ x =9.1 m Aim:

Work (W ) Solution:

W =F Δ xcosθ Since 41 ° is below the horizontal direction then:

θ=360° −41 °=319 ° W =( 45 N ) ( 9.1m ) cos ( 319 °)

W =309 J 3. A constant force of 25 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force?

θ

Given:

F=25 N Δ x =7.5 m θ=180−30 =150° Aim:

Work (W ) Solution:

W =F Δ xcosθ

W =( 25 N )( 7.5 m ) cos(150 °) W =−162 J 4. The kinetic energy of a car is 8 × 10 6 J as it travels along a horizontal road. How much work is required to stop the car in 10 s? Given: 6

K=8 x 10 J

t=10 s Aim:

Work (W ) Solution: Work and kinetic energy are equal. So, the amount of work required to stop the car in 10 s is just the same amount of kinetic energy given in the problem.

W =∆ K

W =8 x 106 J 5. How much energy is dissipated in breaking a 1000-kg car to a stop from an initial speed of 20 m/s? Given:

m=1000 kg v i=20 m/ s Aim:

Kinetic Energy ( K ) Solution:

1 K= m v 2 2

( )

1 m K= ( 1000 kg ) 20 2 s

2

K=200000 J∨2.00 x 105 J 6. The kinetic energy of an 1100-kg truck is 4.6 × 105 J. What is the speed of the truck? Given:

m=1100 kg 5

K=4.6 x 10 J Aim:

velocity (v )

Solution:

1 K= m v 2 2

v 2=

(2) K m

√

( 2) K m

v=

√

5

( 2) (4.6 x 10 J ) v= (1100 kg) v =28. 9

m s

7. A 40-kg block is lifted vertically 20 meters from the surface of the earth. To one significant figure, what is the change in the gravitational potential energy of the block? Given:

m=40 kg h=20 m Aim:

Gravitational Potential Energy (U) Solution:

U= w ´ h=m ´g h

( ms )( 20 m)

U=( 40 kg) 9.8

2

U =7840 J (makeit one sig fig) 3

U =8000∨8 x 1 0 J 8. The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much power is required to stop the car in 10 s? Given: 6

K=8 x 10 J Δ t =10 s Aim:

Power ( P ) Solution:

P=

W ΔK = Δt Δ t

P=

8 x 106 J 10 s 5

P=860000 J∨8.60 x 10 J...