A box starts out at the top of a frictionless ramp, then slides down. The ramp h
ID: 2183314 • Letter: A
Question
A box starts out at the top of a frictionless ramp, then slides down. The ramp has a heighth=3meters and a slope of5degrees with respect to the ground. The box has a mass of3kg. The system is under normal gravity, sog= 9.81 m/s2Let the potential energy be zero at the bottom.
a) What is the initial gravitational energy of the box at the top of the ramp?
J
b) What is the kinetic energy of the box at the bottom of the ramp?
J
c) What is the total mechanical energy of the box at the bottom of the ramp?
J
d) What is the total mechanical energy of the box at the top of the ramp?
J
e) What does this make the velocity of the box at the bottom?
m/s
f) What is this velocity if the box starts with a mass of9kg?
m/s
Explanation / Answer
A 3 kg box slides down a long, frictionlessincline of angle 35. It starts from rest at time
t = 0 at the top of the incline at a height 35 m
above the ground.
The acceleration of gravity is 9.81 m/s2 .
a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
c) Find the box’s speed at t = 1.9 s.
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J. A 3 kg box slides down a long, frictionless
incline of angle 35. It starts from rest at time
t = 0 at the top of the incline at a height 35 m
above the ground.
The acceleration of gravity is 9.81 m/s2 .
a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
c) Find the box’s speed at t = 1.9 s.
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J. A 3 kg box slides down a long, frictionless
incline of angle 35. It starts from rest at time
t = 0 at the top of the incline at a height 35 m
above the ground.
The acceleration of gravity is 9.81 m/s2 .
a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
c) Find the box’s speed at t = 1.9 s.
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J. a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
PE = mgh = 3 x 9.81 x 35 = 1030.05 J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
9.81sin(35) = 5.6268 m/s^2
0.5 x 5.6268 x 1.9^2 = 10.1563 m
c) Find the box’s speed at t = 1.9 s.
v = at = 5.6268 x 1.9 = 10.69 m/s
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
0.5 x 3 x 10.69^2 = 171.4141 J
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J.
v = 2gh
(2 x 9.81 x 35) = 26.2 m/s a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
PE = mgh = 3 x 9.81 x 35 = 1030.05 J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
9.81sin(35) = 5.6268 m/s^2
0.5 x 5.6268 x 1.9^2 = 10.1563 m
c) Find the box’s speed at t = 1.9 s.
v = at = 5.6268 x 1.9 = 10.69 m/s
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
0.5 x 3 x 10.69^2 = 171.4141 J
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J.
v = 2gh
(2 x 9.81 x 35) = 26.2 m/s a) What is the original potential energy of
the box relative to the ground?
Answer in units of J
PE = mgh = 3 x 9.81 x 35 = 1030.05 J
b) Find the distance the box travels during
the interval 0 < t < 1.9 s .
Answer in units of m.
9.81sin(35) = 5.6268 m/s^2
0.5 x 5.6268 x 1.9^2 = 10.1563 m
c) Find the box’s speed at t = 1.9 s.
v = at = 5.6268 x 1.9 = 10.69 m/s
d) Find the potential energy of the box at
t = 1.9 s .
Answer in units of J.
0.5 x 3 x 10.69^2 = 171.4141 J
e) Find the speed of the box just as it
reaches the bottom of the incline.
Answer in units of J.
v = 2gh
(2 x 9.81 x 35) = 26.2 m/s
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