A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box
and the ramp.
1 answer:
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
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Answer:
24,187.04 J ≈ 24,200 J
Explanation:
mass (m) = 544 kg
initial speed (u) = 6.75 m/s
final speed (v) = 15.2 m/s
change in height (Δh) = -14 m (negative sign is because there is a decrease in height )
acceleration due to gravity (g) = 9.8 m/s^{2}
How much work was done on the raft by non conservative forces?
work done = change in energy of the system = change in kinetic energy + change in potential energy
work done = () + (mgΔh)
work done = () + (544 x 9.8 x (-14))
work done = 50449.76 - 74,636.8
work done = 24,187.04 J ≈ 24,200 J