A 23 kg child sits at the top of a slide, inclined at 38 degrees, holding onto a

Question

A 23 kg child sits at the top of a slide, inclined at 38 degrees, holding onto a rope with a tension of 10N. If the child accelerates down the slide at 5 m/s2, What is the coefficient of kinetic friction between her and the slide? What coefficient of static friction value(s) between the child and the slide would be capable of keeping her at rest, even with the rope providing 10N of tension?

The correct answers, respectively, are: a) .185 and b) .85 (or greater) Could someone explain how to get these answers?

Explanation / Answer

mass of the child m=23 kg

angle of iclination, theta=38 degrees,

tension T=10 N

acceleration down ward,a=5 m/sec^2

here,

Normal force N=m*g*cos(theta)

frictional force, f=u*N

frictional force, f=u*(m*g*cos(theta))

componenet of gravitational force , F=m*g*sin(thet)

a)

if the body is moving downward

m*a=m*g*sin(theta)+T-f

m*a=m*g*sin(theta)-uk*(m*g*cos(theta))+T

m*a=m*g*(sin(theta)-uk*cos(theta))+T

23*5=23*9.8*(sin(38)-uk*cos(38))+10

===> uk=0.19

coefficient of kinetic friction uk=0.185

b)

if mass is at rest,

m*a=m*g*sin(theta)+T-f

0=m*g*sin(theta)-us*(m*g*cos(theta))+T

0=m*g*(sin(theta)-us*cos(theta))+T

0=23*9.8*(sin(38)-us*cos(38))+10

us=0.84

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