Question

A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if the child given that the coefficent of kinetic friction between the child and the slide is 0.415.

Answer 1

4.16m/s²

Step-by-step explanation:

According to Newtons second law;

`\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\`

Fm is the moving force

`\mu` is the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is 4.16m/s²