Please explain the process I would take to solving this equation and provide a list of formulas used.
A 75 kg box slides down a 25 degree ramp with an acceleration of 3.6m/s/s.
a. Find the coefficient of friction.
b. What acceleration would a 175 kg box have on this ramp?
A 75 kg box slides down a 25 degree ramp with an acceleration of 3.6m/s/s.
a. Find the coefficient of friction.
b. What acceleration would a 175 kg box have on this ramp?
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The process involves sketching the situation, first of all.
From this sketch the acceleration of gravity can be decomposed into
an acceleration down the ramp = g sin 25 = (9.8)(sin 25) = 4.14 m/s²
and one Normal to the ramp = g cos 25 = (9.8)(cos 25) = 8.88 m/s²
since the NET acceleration down the ramp = 3.6 m/s² that means
the decrease in acceleration down ramp of 4.14 - 3.6 = 0.54 m/s² is
caused by friction. The coefficient of kinetic friction = 0.54/8.88 = 0.06 <= ANS (a)
IF the box was 175 kg instead of 75 kg the acceleration of gravity wouldn't change
and neither does the coefficient of kinetic friction - so the NET acceleration down ramp
wouldn't change and be equal to 3.6 m/s² <= ANS (b)
From this sketch the acceleration of gravity can be decomposed into
an acceleration down the ramp = g sin 25 = (9.8)(sin 25) = 4.14 m/s²
and one Normal to the ramp = g cos 25 = (9.8)(cos 25) = 8.88 m/s²
since the NET acceleration down the ramp = 3.6 m/s² that means
the decrease in acceleration down ramp of 4.14 - 3.6 = 0.54 m/s² is
caused by friction. The coefficient of kinetic friction = 0.54/8.88 = 0.06 <= ANS (a)
IF the box was 175 kg instead of 75 kg the acceleration of gravity wouldn't change
and neither does the coefficient of kinetic friction - so the NET acceleration down ramp
wouldn't change and be equal to 3.6 m/s² <= ANS (b)
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a) sum of the forces = ma
Fn - Fk = ma
Fn = normal force = mg
Fk = force of friction = mg * coefficient of friction
The normal force is perpendicular to the incline. So, you need to multiply by cos(theta) in order to get the component of the force along the incline.
(75 kg)(9.8 m/s^2)cos(25) - (75 kg)(9.8 m/s^2) * coefficient of friction = (75 kg)(3.6 m/s^2)
coefficient of friction = 0.54
b) The acceleration is dependent on the coefficient of friction, not the weight of the object. So, it will have the same acceleration as the lighter box.
Fn * cos(theta) - Fk = ma
(175 kg)(9.8 m/s^2)cos(25) - (175 kg)(9.8 m/s^2) * 0.54 = (175 kg)a
a = 3.6 m/s^2
Fn - Fk = ma
Fn = normal force = mg
Fk = force of friction = mg * coefficient of friction
The normal force is perpendicular to the incline. So, you need to multiply by cos(theta) in order to get the component of the force along the incline.
(75 kg)(9.8 m/s^2)cos(25) - (75 kg)(9.8 m/s^2) * coefficient of friction = (75 kg)(3.6 m/s^2)
coefficient of friction = 0.54
b) The acceleration is dependent on the coefficient of friction, not the weight of the object. So, it will have the same acceleration as the lighter box.
Fn * cos(theta) - Fk = ma
(175 kg)(9.8 m/s^2)cos(25) - (175 kg)(9.8 m/s^2) * 0.54 = (175 kg)a
a = 3.6 m/s^2