Question

A child whose weight is 270 N slidesdown a 6.2 m playground slide that makesan angle of 20° with the horizontal. The coefficient of kineticfriction is 0.10.

(a) How much energy is transferred to thethermal energy?
1 J

(b) If she starts at the top with a speed of 4.97 m/s, what is herspeed at the bottom?
2 m/s

Answer 1

Consider



W=mg= 270N
f= μmgcos(20)
(a) How much energy is transferred to the thermal energy?
W _thermal = fL= μmgcos(20) L
W _thermal = fL= 0.10 x 270 cos(20) x 6.3
W _thermal = 160 J

(b) If she starts at the top with a speed of 4.97 m/s, what is herspeed at the bottom?

Pe₁ + Ke₁-W_f=Ke_f  ( the loss due to friction was not figuredin)
mgh + 0.5mV₁² -μmgcos(20) L =0.5mV_f²

V_f² = 2gh +  V₁² - μgcos(20) L
V_f = √(2gh +   V₁²- μgcos(20) L) whereh = L sin(20)
V_f = √(2g L sin(20) +  V₁²    -μgcos(20) L)
V_f = √(2x 9.81 x 6.3 sin(20)+   4.97²- 0.1 x9.81 cos(20) 6.3)
V_f = 7.8m/s

Answer 2

ANSWER :

1. 

(a)

Normal reaction = mg cos(20) = 270 cos(20)

Frictional force = µ * 270 cos(20) = 0.10 * 270 cos(20)

W. D against friction = 27 cos(20) * 6.2   (J)

Energy transferred to thermal energy 

= WD against friction 

= 27 cos(20) * 6.2

= 157.305 J (ANSWER).

(b)

K. E at the top = 1/2 m v^2 = 1/2 W/g v^2 = 1/2 * 270/9.8 * (4.7)^2 = 304.30 J

Height of slide top = 6.2 sin(20) = 2.12 m

P. E at the top of slide = W * h = 270 * 2.12 = 572.4 J

K. E. At the bottom 

= K. E at the top + P.E. at the top - Thermal energy generated

= 304.30 + 572.40 - 157.305

= 719.395 J

Let her speed at the bottom be V m/sec.

So,

1/2 W/g V^2 = 719.395

=> V^2 = 719.395* 2g / W = 719.397 * 2*9.8 / 270 = 52.223

=> V = sqrt(52.223)

=> V = 7.227 m/sec 

Her speed at the bottom of slide is = 7.223 m/sec (ANSWER).