If the crate is given a slight push down the ramp, initially with a speed of 0.75 m/s, what will be (in m/s) right before it reaches the bottom? Hint m/s
If the crate is given a slight push down the ramp, initially with a speed of 0.75 m/s, what will be (in m/s) right before it reaches the bottom? Hint m/s
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Transcribed Image Text:If the crate is given a slight push down the ramp, initially with a speed of 0.75 m/s, what will be the final speed
(in m/s) right before it reaches the bottom?
Hint
m/s
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Transcribed Image Text:Solve for v
Crate Sliding Down a Ramp
A 4.50 kg crate slides down a ramp. The ramp is 1.80 m in length and inclined at an angle of 25.0° as shown in
the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 4.50 N, and
continues to move a short distance on the horizontal floor after it leaves the ramp.
(1) vf =
A crate slides down a ramp under the influence of gravity.
The potential energy of the system decreases, whereas the
kinetic energy increases.
V = 0
Substitute numerical values to calculate the speed (in m/s):
V, =
m/s
(b) How far does the crate slide on the horizontal floor if it continues to experience a friction force of
d
magnitude 4.50 N?
Analyze This part of the problem is handled in exactly the same way as part (a), but in this case we can
consider the mechanical energy of the system to consist only of kinetic energy because the potential
energy of the system ---Select---
v.
(Use the following as necessary: g, f, v, and m.)
(a) Use energy methods to determine the speed of the crate at the bottom of the ramp.
Write the conservation of energy equation for this situation:
SOLUTION
ΔΚ+ ΔΕ 0
Substitute for the energies:
Conceptualize Imagine the crate sliding down the ramp in the figure. The larger the friction force, the
---Select--- v the crate will slide.
0-
Categorize We identify the crate, the surface, and the Earth as an isolated system with a
---Select---
v force acting.
Analyze Because v, = 0, the initial kinetic energy of the system when the crate is at the top of the ramp
is zero. If the y coordinate (in m) is measured from the bottom of the ramp (the final position of the
crate, for which we choose the gravitational potential energy of the system to be zero) with the upward
direction being positive, then y, =
0 = p*j +
m.
Solve for the distance
and substitute numerical values to calculate its numerical value (in m):
(Use the following as necessary: f, m, d, g, and y.)
mv?
d = =
Write the conservation of energy equation for this system:
2f,
ΔΚ + ΔυU + ΔΕ, -0
Finalize For comparison, you may want to calculate the speed of the crate at the bottom of the ramp in
the case in which the ramp is frictionless. Also notice that the increase in internal energy of the system
J. This energy is shared between the crate
Substitute for the energies:
(in J) as the crate slides down the ramp is fd =
and the surface, each of which is a bit warmer than before.
Gm² - 0) • (0 -
Also notice that the distance d the object slides on the horizontal surface is infinite if the surface is
frictionless. Is that consistent with your conceptualization of the situation?
EXERCISE
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