A 15 kg box slides down a long, frictionless incline of angle30°. It starts from

Question

A 15 kg box slides down a long, frictionless incline of angle30°. It starts from rest at time t = 0 at the top ofthe incline at a height of 23 m above ground.

(a) What is the original potential energy of the box relative tothe 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) PE = mgh = 15 * 9.8 * 23 = 3381J b)    The distance traveled by the box            s = ut+ 0.5at2              = 0 + 0.5 gsin t2              = 0.5 *9.8 sin30 * 1              = 2.45 m     speed of the box          v = u + at          v = 0 + gsint       ==> v = 9.8 sin30 * 1                 = 4.9m/s c) KE = 0.5mv2 = 0.5 *15 *4.9 * 4.9 = 180.075J    PE at t = 1s PE1 =PE - KE = 338 - 180.075 = 157.92 J d) At the bottom of the inclined plane the total potentialenergy is converted to KE      ==> KE = 338 J    speed of the box v = 2KE/m                                 = 2*338 / 15                                 = 6.713 m/s

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