60 days, tell me if I'm correct please.
<em>Given that:</em>
mass of the ball (m) = 0.5 Kg ,
ball strikes the wall (v₁) = 5 m/s ,
rebounds in opposite direction (v₂) = 2 m/s,
time duration (t) = 0.01 s,
<em> Determine the force (F) = ?</em>
We know that from Newton's II law,
<em>F = m. a</em> Newtons
(velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>
= m × (v₁ + v₂)/t
= 0.5 × (5 + 2)/0.01
= 350 N
<em>The force acting up on the ball is 350 N</em>
C., used in power plants I think.
weight = mg acts
downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is
acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left
Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ
We let the value of µs be equal to a value such that any F
will not be able to move the crate. Then, if we increase F by an amount F',
then the force pushing the crate towards the right also increases by F' cosΘ. Additionally,
the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase
by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds
by F' sin Θ.
<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>
