A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
.490 T. If the coil carries a current of 26.7 mA, find the magnitude of the maximum possible torque exerted on the coil.
2 answers:
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(a) 69.3 J
The work done by the applied force is given by:
where:
F = 25.0 N is the magnitude of the applied force
d = 3.20 m is the displacement of the sled
is the angle between the direction of the force and the displacement of the sled
Substituting numbers into the formula, we find
(b) 0
The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.
(c) 69.3 J
According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:
where
is the change in kinetic energy
W is the work done
Since we already calculated W in part (a):
W = 69.3 J
We therefore know that the change in kinetic energy of the sled is equal to this value:
(d) 4.9 m/s
The change in kinetic energy of the sled can be rewritten as:
(1)
where
Kf is the final kinetic energy
Ki is the initial kinetic energy
m = 5.50 kg is the mass of the sled
u = 0 is the initial speed of the sled
v = ? is the final speed of the sled
We can calculate the variation of kinetic energy of the sled, , after it has travelled for d=3 m. Using the work-energy theorem again, we find
And substituting into (1) and re-arrangin the equation, we find
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :
So, the crate's coefficient of kinetic friction on the floor is 0.23.