A bullet of mass 0.02070 �� collides inelastically with a wooden block of mass 28.00 ��, initially at rest. after the collision
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Answer:
Minimum coefficient of kinetic friction between the surface and the block is .
Explanation:
Given:
Mass of the block = M
Spring constant = k
Distance pulled = x
According to the question:
<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed. </em>
So,
From the FBD we can say that:
⇒ Normal force, <em>...equation(i)</em>
⇒ Elastic potential energy, =
<em> ...equation (ii)</em>
⇒ Frictional force, =
<em> ...equation (iii)</em>
⇒ Plugging (i) in (iii).
⇒
Now,
⇒ As we know that the energy lost due to friction is equivalent to PE .
⇒ <em>...considering PE as</em>
or
.
Arranging the equation.
⇒
⇒ <em>...eliminating x from both sides.</em>
⇒ <em>...dividing both sides wit Mg.</em>
Minimum coefficient of kinetic friction between the surface and the block is .
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.