Asked by Anonymous
A 5-kg container slides down an incline at an angle of 28 degrees. If the coefficient of kinetic friction is 0.4, what is the container's acceleration when it slides down the incline, and how long does it take for the container to travel 14 meters? (Use seconds as the unit of time.)
My answer:
The acceleration is 4.6 meters/sec^2, and it takes 2.5 seconds to travel the distance (I used the kinematic equations).
My answer:
The acceleration is 4.6 meters/sec^2, and it takes 2.5 seconds to travel the distance (I used the kinematic equations).
Answers
Answered by
Henry
M*g = 5 * 9.8 = 49 N. = Wt. of container.
Fp = 49*sin28 = 23.0 N. = Force parallel to the incline.
Fn = 49*Cos28 = 43.3 N. = Normal force.
Fk = u*Fn = 0.4 * 43.3 = 17.3 N. = Force of kinetic friction.
a. Fp-Fk = M*a.
23-17.3 = 5*a,
a = 1.14 m/s^2.
b. d = 0.5a*t^2.
t = ?.
Fp = 49*sin28 = 23.0 N. = Force parallel to the incline.
Fn = 49*Cos28 = 43.3 N. = Normal force.
Fk = u*Fn = 0.4 * 43.3 = 17.3 N. = Force of kinetic friction.
a. Fp-Fk = M*a.
23-17.3 = 5*a,
a = 1.14 m/s^2.
b. d = 0.5a*t^2.
t = ?.
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