The momentum of an object is 35 kg•m/s and it is travelling at a speed of 10 m/s.
1 answer:
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Answer:
(a) the mechanical energy of the system, U = 0.1078 J
(b) the maximum speed of the object, Vmax = 0.657 m/s
(c) the maximum acceleration of the object, a_max = 15.4 m/s²
Explanation:
Given;
Amplitude of the spring, A = 2.8 cm = 0.028 m
Spring constant, K = 275 N/m
Mass of object, m = 0.5 kg
(a) the mechanical energy of the system
This is the potential energy of the system, U = ¹/₂KA²
U = ¹/₂ (275)(0.028)²
U = 0.1078 J
(b) the maximum speed of the object
(c) the maximum acceleration of the object
Answer:
Work done = 422.45 kJ
Explanation:
given,
weight of equipment = 6 kN
coefficient of kinetic friction = 0.05
distance up to which it is pulled = 1000 m
constant acceleration = 0.2 m/s²
Work done by the camper = ?
actual acceleration acting a'
m a = m a' - μ mg
a' = a + μ g
a' = 0.2 + 0.05 x 9.8
a' = 0.69 m/s²
Work done = Force x distance
F = m a'
F = 422.44897 N
Work done = F x d
Work done = 422.44897 x 1000
Work done = 422449 J
Work done = 422.45 kJ