Question

A child whose weight is 276 N slides down a 5.90 m playground slide that makes an angle of 34.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.130. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.518 m/s, what is her speed at the bottom?

Answer 1

Answer:

Explanation:

weight of child, mg = 276 N

distance, s = 5.90 m

angle of inclination, θ = 34°

coefficient of kinetic friction, μk = 0.130

(a)

The work done by the friction force is converted into heat.

W = μk x mg x Cos θ x s

W = 0.130 x 276 x Cos 34° x 5.90

W = 175.5 Joule

Thus, the amount of thermal energy is 175.5 Joule.

(b)

initial speed, u = 0.518 m/s

Let the speed at the bottom is v.

let the acceleration is a.

a = g Sinθ - μk g Cosθ

a = 9.8 (Sin 34 - 0.130 x Cos 34)

a = 9.8 (0.559 - 0.108)

a = 4.42 m/s

Use III equation of motion

v² = u² + 2 a s

v² = 0.518 x 0.518 + 2 x 4.42 x 5.9

v² = 0.2683 + 52.156

v = 7.24 m/s

Answer 2

Answer:

Explanation:

Frictional force acting on the child = μ mg cosθ

, μ is coefficient of kinetic friction ,  m is mass of child θ is inclination

work done by frictional force

μ mg cosθ x d , d is displacement on inclined plane

work done = .13 x 276 x cos34 x 5.9

= 175.5 J

This work will be converted into heat energy.

b ) Initial energy of child = mgh + 1/2 m v ² , h is height , v is initial velocity

= 276 x 5.9 sin34  + 1/2 x 276 / 9.8 x .518² [ mass m = 276 / g ]

= 910.59 + 3.77

= 914.36 J

loss of energy due to friction = 175.5

Net energy at the bottom

= 738.86 J

If v be the velocity at the bottom

1/2 m v² = 738 .86

.5 x (276 / 9.8) x v² = 738.86

v² = 52.47

v = 7.24 m /s .