A 10 kg box slides down a long, frictionless incline of angle 30°. It starts fro
ID: 1353958 • Letter: A
Question
A 10 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 21 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)
Original potential energy = mg * h
Original potential energy = 10 * 9.8 * 21
Original potential energy = 2058 J
the Original potential energy is 2058 J
b)
acceleration of block ,a = g * sin(theta)
a = 9.8 * sin(30)
a = 4.9 m/s^2
for distance travalled in 1 s
d = 0.5 at^2
d = 0.5 * 4.9 *1^2
d = 2.45 m
distance travelled is 2.45 m
v = a * t
v = 4.9 * 1 = 4.9 m/s
the speed is 4.9 m/s
c)
at this position
KInetic energy = 0.5 mv^2
KInetic energy = 0.5 * 7 * 4.9^2
KInetic energy = 84 J
potential energy = initial energy - KInetic energy
potential energy = 2058 - 84
potential energy =1974 J
d)
at the bottom of incline
KInetic energy = KInetic potential energy
KInetic energy = 2058 J
0.5 * 7 * v^2 = 2058
solving for v
v = 24.2 m/s
the speed of box is 24.2 m/s
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