Energy -- Review Problems

Solutions to Problems
Irwin Physics Book 2 - Concepts & Connections: Chapter 5

1. a) W = FDd = (40 N)(0.15 m)

W = 6.0 J

b) W = FDd = mgDd = (50 kg)(9.8 N/kg)(1.95 m)

W = 9.6 X 10² J

c) W =  FDd cos Q= (120 N)(4 m)(cos 25°) =  4.4 X 10² J

2. v = 45 km/h = 12.5 m/s

To find Dd, use v₂²  = v₁² + 2aDd  and solve for Dd

Dd = 31.25 m

W = FDd = (5000 N)(31.25 m) = 1.6 X 10² J

3. W = FDd cos Q = (78 N)(10 m)(cos 55°)= 4.5 X 10² J

4. To find a use a = (v₂- v₁)/Dt

a = -2.2 m/s²

F = (52 000 kg)(-2.2 m/s²) = -114 400 N

To find Dd, use v₂²  = v₁² + 2aDd  and solve for Dd

Dd = 97.5 m

W = FDd =  (-114 400 N)(97.5 m) = -1.1 X 10⁷J

5. a) W = FDd = (175 N)(55 m) =  9625 J

b) The triangular areas above and below the axis are identical and cancel out, therefore,

W = (0.040 m)(20 N) = 0.80 J

6. F = ma =  (3 kg)(9.8 N/kg) = 29.4 N

From W = FDd  calculate Dd = W / F = 16 m

1. a) Ek = (mv²)/2 = (20 000 kg)(7500 m/s)²/2  = 5.6 X 10¹¹J

b) v =20 km/h  =  5.6 m/s

c) Ek = (mv²)/2 = (0.030 kg)(400 m/s)²/2  = 2.4 X 10³J

v = 5.6 m/s

m = 6.5 kg

4.

p = 4.2 X 10^-23 N^.s

More Solutions

Related Topics:

Momentum
Impulse
Conservation Of Momentum