A box is sliding up an incline that makes an angle of with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.40 m/s. How far does the box travel along the incline before coming to rest

0.231 m

Explanation:

Given that

Initial velocity, u = 1.4 m/s

Coefficient if friction, μ = 0.180

Final velocity, v = 0 m/s

Since the angle made with the inclined isn't stated. I'd be working with an angle of 15°.

The formula for the acceleration is given as

Acceleration = f.r.m.g.sinθ/m

Where f = μN = μ.m.g.cosθ

On substituting, acceleration a then is equal to

a = -(u.g.cosθ + g.sinθ)

a = -(0.18 * 9.8 * cos15 + 9.8 sin15)

a = -(1.764 * 0.9659 + 9.8 * 0.2588)

a = -(1.7038 + 2.5362)

a = -4.24 m/s²

To find the distance traveled, we use one of the equations of motions

v² = u² + 2as, with v = 0

u² = -2as

s = -u²/2a, on substituting

s = -1.4² / -2 * 4.24

s = 1.96 / 8.48

s = 0.231 m

i think they are all correct

explanation:

they seem good

double the distance = divide strength by 4