A terrier has been trained to slide down a ramp of length 10 m. First the dog cl

Question

A terrier has been trained to slide down a ramp of length 10 m. First the dog climbs vertically 8m then slides down in 1.6 seconds. The local value of g is 9.8 m/s^2. What is the coefficient of friction that exists between the dogs feet and the ramp. Treat the dog as a point object and use only one frictional force.

B) what is the terriers acceleration using kinematics

C) using newtons second law and free body diagram solve x and y axis equations to get coefficient of friction of the dog coming down the ramp.

Explanation / Answer

a)

imagine the ramp as a right angled triangle whose height is 8 m, hypotenus is 10 m

then base will be sqrt(10^2-8^2)=6 m

angle with horizontal=arctan(8/6)=53.13 degrees

then normal force =mass*g*cos(53.13)=5.88*m N

where m=mass of the dog.

if friction coefficient is k, then friction force=k*normal force=5.88*m*k, opposing the motion

component of dog's weight, acting along the direction of motion=m*9.8*sin(53.13)=7.84*m N

then net force=7.84*m-5.88*m*k

net acceleration=force/mass=7.84-5.88*k

now as it slides down 10 m in 1.6 seconds, if acceleration is a

then 0.5*a*1.6^2=10

==>a=7.8125 m/s^2

==>7.84-5.88*k=7.8125

==>k=4.67*10^(-3)

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