A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction
1 answer:
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Answer:
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
Eq (2)
Divide Eq(2) and Eq (1)
The Required horizontal force is 230.04N