I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
Answer:
Vmax=11.53 m/s
Explanation:
from conservation of energy
Spring potential energy =potential energy due to elevation
0.5*k*x²= mg
=mgh
0.5*k*2.3²= 430*9.81*6
k=9568.92 N/m
For safety reason
k"=1.13 *k= 1.13*9568.92
k"=10812.88 N/m
agsin from conservation of energy
spring potential energy=change in kinetic energy
0.5*k"*x²=0.5*m*
10812.88 *2.3²=430*
=11.53 m/s
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