Question

A child whose weight is 267 N slides down a 6.10 m long slide that makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction between the slide and the child is 0.100. If the child starts at the top with a speed of 0.457 m/s, what is the child's speed at the bottom? (Ignore air resistance)

Answer 1

Here we have given that,

W = 267 N

Slide length L = 6.1 m

Angle = 20°

Uk = 0.1

Initial speed of child u = 0.457 m/s

Negligible air resistance.

Here the vertical height of the child will be given as,

sinθ = h/6.1

h = 6.1 x sin20°

h = 2.086 m

Here energy got wasted in the heat form by frictional force will be given as

W = ukWCos20°×6.1 =(153.0477371474) J

Now for the velocity of the child at bottom we have to apply here the law of energy conservation which is given as,

KE(bottom) = KE(top) + PE(top) - W (friction)

1/2mv²= 1/2mu²+ mgh - W

1/2 x (267/9.8) x v²= 1/2 x (267/9.8) x (0.457)^2 + 267 x 2.086 - 153.04773 J

v = = 5.4649 m/s

Hence the childs speed at the bottom will be 5.4649 m/s