A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coefficient of static friction between the car's tires and the road is 0.300. What is the magnitude of the force exerted by friction on the car?

1593.4 N

Explanation:

The road is banked at an angle of 25°. The frictional force along the banked road f. The horizontal component of the car's weight along the road is F = mgsinθ. The horizontal net force along the road is thus F₁ = F - f. F₁ also equals the centripetal force on the car as it goes round the curve. So, F₁ = mv²cosθ/r

F - f = mv²cosθ/r

mgsinθ - f = mv²cosθ/r

f = mgsinθ - mv²cosθ/r

f = m(gsinθ - v²cosθ/r)

Given m = mass of car = 600 kg, g = 9.8 m/s², θ = banking angle = 25°, v = speed of car = 30 m/s and r = radius of curved road = 120 m

f = 600(9.8sin25 - 30²cos25/120)

f = 600(4.1417 - 6.7973)

f = 600(-2.6556)

f = -1593.4 N

This is the friction between the road and the tyres. The magnitude of the frictional force exerted on the car is 1593.4 N

Since f = μmgcosθ where  μ = coefficient of static friction = 0.3

f = 0.3 × 600 × 9.8cos25 = 1598.73 N

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