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WCU PHYS 261 MIDTERM EXAM REVIEW 2026–2027:
PRACTICE QUESTIONS WITH 100% CORRECT ANSWERS –
HUMAN PHYSIOLOGY (WEST COAST UNIVERSITY /
NURSING & PRE‑MED / ALLIED HEALTH)
Work on a Block Sliding Up a Frictionless Incline: ......ANSWER......Block
weight 15.0 N sits on a frictionless inclined plane, which makes an angle
θ = 23.0° with respect to the horizontal, as shown in the figure. A force
of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient
to pull the block up the plane at constant speed.
a) The block moves up an incline with constant speed. What is the total
work W. Total done on the block by all forces as the block moves a
distance L = 3.40 m up the incline? Include only the work done after the
block has started moving at constant speed, not the work needed to
start the block moving from rest. ......ANSWER......Since the block is
moving at a constant speed there is no change in kinetic energy. Since
work is the change in kinetic energy and there is no change, the net
work on the block is zero.
b) What is Wg, the work done on the block by the force of gravity w‖ as
the block moves a distance L = 4.50m up the incline? ......ANSWER......1)
Find the component of the gravitational force parallel to the plane
What is w||, the magnitude of the component of the force of gravity
along the inclined plane? →w parallel to the inclined plane has
magnitude given by w‖=wsinθ. = 40*sin(22) = 15.
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2) Wg = -mgy = -15*4.5 = -67J
What is Wƒ, the work done on the block by the applied force F→ as the
block moves a distance = 4.5m up the incline? ......ANSWER......Since we
found the net work to be zero (in Part A), the work done by the applied
force has to offset the work done by gravity: Wg=−Wƒ. ∴ 67J
The mechanical energy of a system is defined as the sum of kinetic
energy K and potential energy U. For such systems in which no forces
other than the gravitational and elastic forces do work, the law of
conservation of energy can be written as ......ANSWER......Ki+Ui=Kf+Uf,
K=½mv²
Potential: Ug = mgh
Elastic potential energy: For a spring with a force constant k, stretched
or compressed a distance x, the associated elastic potential energy is
Ue=½kx²
b) Which form of the law of conservation of energy describes the
motion of the block when it slides from the top of the table to the
bottom of the ramp? ......ANSWER......Think about these questions:
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Are there any nonconservative forces acting on the block during this
part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
½mv₁² + mgh₁ = ½mv₂² + mgh₂
c) What if nonconservative forces, such as friction, also act within the
system? In that case, the total mechanical energy will change. The law
of conservation of energy is then written as ......ANSWER......½mv₁² +
mgh₁ + ½kx₁² + Wnc = ½mv₂² + mgh₂ + ½kx₂²
Where Wnc represents the work done by the nonconservative forces
acting on the object between the initial and the final moments. The
work Wnc is usually negative; that is, the nonconservative forces tend
to decrease, or dissipate, the mechanical energy of the system.
d) Using conservation of energy, find the speed vb of the block at the
bottom of the ramp.
Express your answer in terms of some or all the variables m, v, h and
and any appropriate constants. ......ANSWER......PE₀ + KE₀ = PEƒ + KEƒ
At the top of the ramp, potential energy will be at a maximum and
kinetic energy will be at a minimum. At the bottom of the ramp, kinetic
energy will be at a maximum and potential energy will be minimum. But
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although the type of energy changes, the total amount of energy
remains the same:
mgh + ½mv² = mgh(b) + ½mv(b)²
Since the height at the bottom is zero, we can eliminate potential
energy from the right side of the formula:
mgh + ½mv² = ½mv(b)²
gh + ½v² = ½v(b)²
2gh + v² = v(b)²
v(b) = √(v² + 2gh)
== √(v² + 2gh)
As the block slides across the floor, what happens to its kinetic energy K,
potential energy U, and total mechanical energy E? ......ANSWER......K
decreases
U stays the same
E decreases
g) What force is responsible for the decrease in the MECHANICAL
ENERGY of the block? ......ANSWER......friction
WCU PHYS 261 MIDTERM EXAM REVIEW 2026–2027:
PRACTICE QUESTIONS WITH 100% CORRECT ANSWERS –
HUMAN PHYSIOLOGY (WEST COAST UNIVERSITY /
NURSING & PRE‑MED / ALLIED HEALTH)
Work on a Block Sliding Up a Frictionless Incline: ......ANSWER......Block
weight 15.0 N sits on a frictionless inclined plane, which makes an angle
θ = 23.0° with respect to the horizontal, as shown in the figure. A force
of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient
to pull the block up the plane at constant speed.
a) The block moves up an incline with constant speed. What is the total
work W. Total done on the block by all forces as the block moves a
distance L = 3.40 m up the incline? Include only the work done after the
block has started moving at constant speed, not the work needed to
start the block moving from rest. ......ANSWER......Since the block is
moving at a constant speed there is no change in kinetic energy. Since
work is the change in kinetic energy and there is no change, the net
work on the block is zero.
b) What is Wg, the work done on the block by the force of gravity w‖ as
the block moves a distance L = 4.50m up the incline? ......ANSWER......1)
Find the component of the gravitational force parallel to the plane
What is w||, the magnitude of the component of the force of gravity
along the inclined plane? →w parallel to the inclined plane has
magnitude given by w‖=wsinθ. = 40*sin(22) = 15.
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2) Wg = -mgy = -15*4.5 = -67J
What is Wƒ, the work done on the block by the applied force F→ as the
block moves a distance = 4.5m up the incline? ......ANSWER......Since we
found the net work to be zero (in Part A), the work done by the applied
force has to offset the work done by gravity: Wg=−Wƒ. ∴ 67J
The mechanical energy of a system is defined as the sum of kinetic
energy K and potential energy U. For such systems in which no forces
other than the gravitational and elastic forces do work, the law of
conservation of energy can be written as ......ANSWER......Ki+Ui=Kf+Uf,
K=½mv²
Potential: Ug = mgh
Elastic potential energy: For a spring with a force constant k, stretched
or compressed a distance x, the associated elastic potential energy is
Ue=½kx²
b) Which form of the law of conservation of energy describes the
motion of the block when it slides from the top of the table to the
bottom of the ramp? ......ANSWER......Think about these questions:
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Are there any nonconservative forces acting on the block during this
part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
½mv₁² + mgh₁ = ½mv₂² + mgh₂
c) What if nonconservative forces, such as friction, also act within the
system? In that case, the total mechanical energy will change. The law
of conservation of energy is then written as ......ANSWER......½mv₁² +
mgh₁ + ½kx₁² + Wnc = ½mv₂² + mgh₂ + ½kx₂²
Where Wnc represents the work done by the nonconservative forces
acting on the object between the initial and the final moments. The
work Wnc is usually negative; that is, the nonconservative forces tend
to decrease, or dissipate, the mechanical energy of the system.
d) Using conservation of energy, find the speed vb of the block at the
bottom of the ramp.
Express your answer in terms of some or all the variables m, v, h and
and any appropriate constants. ......ANSWER......PE₀ + KE₀ = PEƒ + KEƒ
At the top of the ramp, potential energy will be at a maximum and
kinetic energy will be at a minimum. At the bottom of the ramp, kinetic
energy will be at a maximum and potential energy will be minimum. But
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although the type of energy changes, the total amount of energy
remains the same:
mgh + ½mv² = mgh(b) + ½mv(b)²
Since the height at the bottom is zero, we can eliminate potential
energy from the right side of the formula:
mgh + ½mv² = ½mv(b)²
gh + ½v² = ½v(b)²
2gh + v² = v(b)²
v(b) = √(v² + 2gh)
== √(v² + 2gh)
As the block slides across the floor, what happens to its kinetic energy K,
potential energy U, and total mechanical energy E? ......ANSWER......K
decreases
U stays the same
E decreases
g) What force is responsible for the decrease in the MECHANICAL
ENERGY of the block? ......ANSWER......friction
