Question

Allison pulls a sled up a hill, which has an incline of 20 degrees to the horizontal. Of the sled has a mass of 20 kg, and the coefficient of kinetic friction between the sled and the ground is 0.1, what is the minimum force Allison must apply to the sled to keep it moving forward at a constant speed

Answer 1

`F=48.62N`

Step-by-step explanation:

From the question we are told that:

Angle
`\theta= 20`

Mass
`m= 20kg`

Coefficient of kinetic friction
`\mu=0.1`

Generally the equation for Force Required to jeep sled moving is mathematically given by

`F= mg sin \theta - \mu N`

Where N is normal force

`F_N=Wcos\theta`

`F_N=20*9.8*cos 20`

`F_N=184.18N`

Therefore

`F= (20*9.8) sin 20 - (0.1) (184.18)`

`F=48.62N`