Jayne lifts the barbell 120 cm upwards. She has a mass of 60kg. How much work does she do?
1 answer:
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Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
Hence, this is the required solution.