A 2.97 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 12.9 N. The coefficient of kinetic friction is 0.126. The acceleration of gravity is 9.8 m/s 2 . Find the speed of the block after it has moved 2.92 m. Answer in units of m/s.

The speed of the block after it has moved 2.92 m is 4.26 m/s

Explanation:

Given;

mass of the block, m = 2.97 kg

initial velocity of the block, u = 0

applied horizontal force, Fx = 12.9 N

Coefficient of static friction, μ = 0.126

The frictional force is given by;

The net horizontal force is given by;

The acceleration of the block is given by;

The speed of the block after it has moved 2.92 m is given by the following kinematic equation;

v² = u² + 2as

v² = 0 + 2as

v² = 2as

v = √2as

v = √(2 x 3.11 x 2.92)

v = 4.26 m/s

Therefore, the speed of the block after it has moved 2.92 m is 4.26 m/s

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

chemical properties of thing