A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?

3.5 m/s

Explanation:

There are 3 forces on the cyclist:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

∑F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

a = -(10 m/s²)(0.60)

a = -6 m/s²

Velocity reached at end of 11 m:

v² = v₀² + 2aΔx

v² = (12 m/s)² + 2 (-6 m/s²) (11 m)

v = √12 m/s

v ≈ 3.5 m/s

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards