Problem 1: (a) Ã= 3â – 2ŷ, B = 4ây – 5ŷ. Ã · B =? Answer: 22 (b) Ã = -2ô + 4, B = 3â – lŷ. Ả.B =? Answer: –10 (c) Ã = 5â – 3ŷ, B = -3â – 5ŷ. Ã. B =? Answer: 0 d = 4m F = 20 N 0=30° 4 kg m = . m = 2 kg HK = 0.5 F = 40 N Hk = 0.2 0 = 30° Problem 2: In the figure above on the left we have a force F = 20 N pulling a box to the right at an angle 0 = 30° above the horizontal. The box has a mass m = 2 kg, the coefficient of kinetic firction between the box and the ground is uk = 0.2, and the block moves a distance d = 4 m to the right. (a) What is the work done by the force F? Answer: 69.3 J (b) What is the normal force from the ground? Use g = 10 m/sec2 Answer: 10 N (c) What is the work done by the kinetic friction force? Answer: -8 J (d) What is Wnet, the net work done on the box? Answer: 61.3 J (e) The box has an initial speed of 4 m/sec. What is the speed of the box after it has travelled the distance d = 4 m? Answer: 8.8 m/sec
Problem 1: (a) Ã= 3â – 2ŷ, B = 4ây – 5ŷ. Ã · B =? Answer: 22 (b) Ã = -2ô + 4, B = 3â – lŷ. Ả.B =? Answer: –10 (c) Ã = 5â – 3ŷ, B = -3â – 5ŷ. Ã. B =? Answer: 0 d = 4m F = 20 N 0=30° 4 kg m = . m = 2 kg HK = 0.5 F = 40 N Hk = 0.2 0 = 30° Problem 2: In the figure above on the left we have a force F = 20 N pulling a box to the right at an angle 0 = 30° above the horizontal. The box has a mass m = 2 kg, the coefficient of kinetic firction between the box and the ground is uk = 0.2, and the block moves a distance d = 4 m to the right. (a) What is the work done by the force F? Answer: 69.3 J (b) What is the normal force from the ground? Use g = 10 m/sec2 Answer: 10 N (c) What is the work done by the kinetic friction force? Answer: -8 J (d) What is Wnet, the net work done on the box? Answer: 61.3 J (e) The box has an initial speed of 4 m/sec. What is the speed of the box after it has travelled the distance d = 4 m? Answer: 8.8 m/sec
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Transcribed Image Text:w = FjAc = F .Aë , W = m(v;)* - m(u,)* , v
1
zm(us)* - m(v,)* , W =
Problem 1: (a) Ã = 3â – 2ŷ, B = 4â& – 5ŷ. Ã ·B =? Answer: 22
(b) Ã = -2â + 4ŷ, B = 3âu – lŷ. Ā·B =? Answer: –10
(c) à = 5â – 3ŷ, B = -3â – 5ŷ. ÷B =? Answer: 0
1
| F,dx , W =
F. dř , F,
dU
W = -AU
dx
d = 4m
F = 20 N
0 = 30°
m = 4 kg
Hk = 0.5)
m = 2 kg
F = 40 N
Hk = 0.2
Ө — 30°
Problem 2: In the figure above on the left we have a force F = 20 N pulling a box to the right at an angle
0 = 30° above the horizontal. The box has a mass m = 2 kg, the coefficient of kinetic firction between the
box and the ground is uk = 0.2, and the block moves a distance d = 4 m to the right.
(a) What is the work done by the force F? Answer: 69.3 J
(b) What is the normal force from the ground? Use g = 10 m/sec² Answer: 10 N
(c) What is the work done by the kinetic friction force? Answer: -8 J
(d) What is Wnet, the net work done on the box? Answer: 61.3 J
(e) The box has an initial speed of 4 m/sec. What is the speed of the box after it has travelled the distance
d = 4 m? Answer: 8.8 m/sec
||
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