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kinetic friction problem?
a kid slides down a 43 degree (from the horizontal) hill with an acceleration of .45 m/s^2 in the opposite direction of motion. what is the coefficient of kinetic friction?
i take it to mean that the frictional force is .45 so:
frictional force = mu * normal force
so mu = frictional force / normal force
and mu = .45 / mgcos(43 degrees)
i may have the normal force wrong im not sure, also mass was not given
help!
thanks!
1 Answer
- DambarudharLv 77 years ago
Mass of the kid = m
Acceleration due to gravity = g
Weight of the kid = w = mg
AB - horizontal plane
AC - inclined surface of the hill
< BAC = Θ = 43º
P - position of the kid at any instant of time
on the inclined surface.
Forces acting on the kid : -
1) Its weight w = mg acting vertically downwards.
Resolve w into 2 components.
(i) w cos Θ, normal to AC downwards
(ii) w sin Θ , parallel to AC downwards.
2) N = normal reaction from AC acting upwards
3) Frictional force f , acting parallel to AC upwards
opposite to motion
Now, N - w cos Θ = 0 ; => N = w cos Θ ------ (1)
f = µ N = µ w cos Θ ---------- (2)
µ = coefficient of friction
Net force acting on the kid = F = w sin Θ - f
= w sin Θ - µ w cos Θ
Applying Newton,s 2nd law of motion, we have
F = w sin Θ - µ w cos Θ = ma --------- (3)
a = acceleration of the kid parallel to AC downward = 4.5 m/s²
(Check the value of 'a' in the question)
Eqn. (3) can be written as : mg sin Θ - µ mg cos Θ = ma
=> g sin Θ - µg cos Θ = a
=> sin Θ - µ cos Θ = a/g
=> µ cos Θ = sin Θ - a/g
=> µ = tan Θ - a/(g cos Θ)
= tan 43 - 4.5/(9.8 cos 43)
= 0.932 - 4.5/(9.8 * 0.731)
= 0.932 - 4.5/7.164
= 0.932 - 0.628 = 0.304 ------ Answer
