Force using Newton's second law

[ruthi]

A child slides down a slide with a 28 degrees incline and at the bottom her speed is preisely half what it would have been if the slide had been firctionless. Caluclate the coefficient of kinetic friction bewteen the slide and the child.

 

[Pyrrhus]

So what have you done?

Hint: Solve for the frictionless case, and for the kinematics consider uniform acceleration.

 

[kyle8921]

Newton's second law states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this scenario, the child's mass remains constant as she slides down the slide. Therefore, the change in her speed is directly proportional to the net force acting on her.

We can use this information to calculate the coefficient of kinetic friction between the slide and the child. The force of friction is equal to the coefficient of friction multiplied by the normal force, which is the force exerted by the slide on the child in a direction perpendicular to the surface. In this case, the normal force is equal to the component of the child's weight that is perpendicular to the slide, which is given by mgcos(28).

The child's final speed is half of what it would have been without friction, so we can set up the following equation:

mgsin(28) - μmgcos(28) = 0.5m(vf^2 - vi^2)

Where m is the mass of the child, g is the acceleration due to gravity, μ is the coefficient of kinetic friction, and vf and vi are the final and initial speeds, respectively.

Solving for μ, we get:

μ = tan(28) - 0.5cot(28)

Plugging in the values, we get μ = 0.267.

Therefore, the coefficient of kinetic friction between the slide and the child is approximately 0.267. This means that the force of friction acting on the child as she slides down the slide is about 26.7% of the component of her weight that is perpendicular to the slide.