A 2.0-kg block sits at the top of an inclined plane with a length of 4.12 m. If the coefficient of kinetic friction μk is 0.18 between the block and the plane, how long will it take the block to slide down the plane when released?

2.16sec

Explanation:

Data;

Mass = 2.0kg, a = 4.12m, u = 0.18, t = ?

F = μN

F = ma and N = mg

ma = μmg

Mass cancels out on booths sides,

Therefore,

μ = a / g

a = μg = 0.18*9.8

a = 1.764m/s²

Using equation of motion,

S = ut + ½at²

But u = 0m/s since the body is at rest

4.12 = 0 + ½ * 1.764 * t²

4.12 = 0.882t²

t² = 4.12 / 0.882

t² = 4.6712

t = √(4.6712)

t = 2.16s

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j