A 13 kg box slides down a long, frictionless incline of angle 30°. It starts fro

Question

A 13 kg box slides down a long, frictionless incline of angle 30°. It starts from rest at time t = 0 at the top of the incline at a height of 16 m above ground.

(a) What is the original potential energy of the box relative to the ground?
______________________J

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

(c) Find the potential energy and the kinetic energy of the box at t = 1 s.
________________J (kinetic energy)
_____________________J (potential energy)

(d) Find the kinetic energy and the speed of the box just as it reaches the bottom of the incline.
________________J
_________________m/s

Explanation / Answer

a)

m = mass = 13 kg

h = height = 16 m

PE = mgh = 13 x 9.8 x 16 = 2038.4 J

b)

acceleration of box parallel to incline = a = g Sin30 = 9.8 Sin30 = 4.9 m/s2

Vo = initial velocity = 0 m/s

t = time taken = 1 sec

distance travelled is given as

d = Vo t + (0.5) a t2

d = 0 x 1 + (0.5) (4.9) (1)2 = 24.5 m

final velocity is given as

Vf = Vo + at = 0 + 4.9 x 1 = 4.9 m/s

c)

hi = initial height = 16 m

final height = hf = hi - distance travelled down = hi - d Sin30 = 16 - 24.5 Sin30 = 3.75 m

Potential energy , PE = mghf = 13 x 9.8 (3.75) = 477.75 J

Kinetic energy = (0.5) m Vf2 = (0.5) (13) (4.9)2 = 156.1 J

d)

at the bottom , h = height = 0

so PE = mgh = 0

Using conservation of energy

KE at bottom = PE at Top = 2038.4 J

(0.5) m V2 = 2038.4

(0.5) (13) V2 = 2038.4

V = 17.71 m/s

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