Abox mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585n. the coefficient of kinetic friction is 0.23 between the box and the surface.
a) find the normal force on the box
b) find the acceleration on the box

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

a. 235.2 N

b. 22 m/s²

Explanation:

a) As there is no displacement in vertical direction, the normal force would be equal to the weight of the box.

N = mg

N = (24 kg) (9.8 m/s²) = 235.2 N

b) Let the acceleration of the box be a such that net force acting on the box is F.

F = ma

F = applied force - force due to friction

Force due to friction = μ N

F = 585 N - (0.23)(235.2)

m a = 530.9

a = 22 m/s²

568.98

explanation:

its math

the total distance traveled by the car is

30km + 49km = 79 km .