A 14 kg box slides down a long,frictionless incline of angle 30°. It starts from

Question

A 14 kg box slides down a long,frictionless incline of angle 30°. It starts from rest at timet = 0 at the top of the incline at a height of22 m above ground. (a) What is the original potential energy ofthe box relative to the ground?
J

(b) From Newton's laws, find the distance the box travels in 1 sand its speed at t = 1 s.
m
m/s

(c) Find the potential energy and the kinetic energy of the box att = 1 s.
J (kinetic energy)
J (potential energy)

(d) Find the kinetic energy and the speed of the box just as itreaches the bottom of the incline.
J
m/s (a) What is the original potential energy ofthe box relative to the ground?
J

(b) From Newton's laws, find the distance the box travels in 1 sand its speed at t = 1 s.
m
m/s

(c) Find the potential energy and the kinetic energy of the box att = 1 s.
J (kinetic energy)
J (potential energy)

(d) Find the kinetic energy and the speed of the box just as itreaches the bottom of the incline.
J
m/s

Explanation / Answer

a) original potential energy PE0 = mgh =14kg*9.8m/s2*22m = 3018.4J b) using Newton's law, we have: F = mgsin(30) = ma =>a = gsin(30) = 4.9m/s2 distance d = (1/2)at2 at t=1.0s , d =0.5*4.9m/s2*(1.0s)2 = 2.45m speed at t = 1.0s is: v(t=1.0s) = at = 4.9m/s2*(1.0s)2 =4.9m/s c) the height to the ground at t=1.0s is: h1 = 22m - 2.45m*sin(30) = 20.775m PE1 = mgh1 = 14kg*9.8m/s2*20.775m = 2850.33J KE1 = (1/2)mv(t=1.0)2 =0.5*14kg*(4.9m/s)2 = 168.07J d) for energy conservation, KEbottom = PE0 = 3018.4J let speed vbottom KEbottom = (1/2)mvbottom2 =3018.4J vbottom = (2*3018.4J/14kg) = 20.77m/s KEbottom = (1/2)mvbottom2 =3018.4J vbottom = (2*3018.4J/14kg) = 20.77m/s

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