Physics for Scientists & Engineers: Study Guide (2026/2027 Syllabus)

Physics for Scientists & Engineers: Study Guide (2026/2027
Syllabus)
This guide is structured unit-by-unit. Each unit contains key concepts, important
equations, and practice questions with detailed answers.




Unit 1: Mechanics - Kinematics & Newton's Laws
Key Concepts: Vectors vs. Scalars, Displacement, Velocity, Acceleration, Uniformly
Accelerated Motion, Projectile Motion, Relative Velocity, Newton's 1st, 2nd, and 3rd
Laws, Free-Body Diagrams, Friction.

Q1: A car accelerates from rest at a constant rate of 2.5 m/s22.5m/s2 for 8.0 seconds.
What is its final velocity and how far does it travel?
ANSWER ✓ The final velocity is 20 m/s20m/s and the distance traveled is 80 m80m.

 Explanation: Using kinematics equations for constant acceleration
(a=2.5 m/s2,t=8.0 s,v0=0a=2.5m/s2,t=8.0s,v0=0).
o Final velocity: v=v0+at=0+(2.5)(8.0)=20 m/sv=v0+at=0+(2.5)(8.0)=20m/s.
o Distance: Δx=v0t+12at2=0+12(2.5)(8.0)2=0.5∗2.5∗64=80 mΔx=v0t+21at2=0+21
(2.5)(8.0)2=0.5∗2.5∗64=80m.
Q2: A ball is thrown vertically upward with an initial velocity of 15 m/s15m/s. How high
does it go? (Ignore air resistance, g=9.8 m/s2g=9.8m/s2)
ANSWER ✓ The maximum height is approximately 11.5 m11.5m.

 Explanation: At the maximum height, the velocity is 0. Using v2=v02+2a(y−y0)v2=v02
+2a(y−y0), where a=−ga=−g, v=0v=0, v0=15 m/sv0=15m/s.
o 0=(15)2+2(−9.8)Δy0=(15)2+2(−9.8)Δy
o 19.6Δy=22519.6Δy=225
o Δy=225/19.6≈11.48 mΔy=225/19.6≈11.48m.
Q3: A projectile is fired at an angle of 30∘30∘ above the horizontal with a speed
of 50 m/s50m/s. What are the horizontal and vertical components of its initial velocity?

, ANSWER ✓ The horizontal component is 43.3 m/s43.3m/s and the vertical component
is 25.0 m/s25.0m/s.

 Explanation:
o v0x=v0cos⁡θ=50cos⁡(30∘)=50∗0.8660≈43.3 m/sv0x=v0
cosθ=50cos(30∘)=50∗0.8660≈43.3m/s.
o v0y=v0sin⁡θ=50sin⁡(30∘)=50∗0.5=25.0 m/sv0y=v0
sinθ=50sin(30∘)=50∗0.5=25.0m/s.
Q4: A 5.0 kg box is pulled across a horizontal floor with a rope that has a tension of 40
N at an angle of 30∘30∘ above the horizontal. If the frictional force is 10 N, what is the
acceleration of the box?
ANSWER ✓ The acceleration of the box is 4.93 m/s24.93m/s2.

 Explanation: First, find the horizontal component of the applied
force: Fx=Tcos⁡θ=40cos⁡(30∘)=34.64 NFx=Tcosθ=40cos(30∘)=34.64N.
The net force in the horizontal direction
is Fnet=Fx−Ffric=34.64 N−10 N=24.64 NFnet=Fx−Ffric=34.64N−10N=24.64N.
Using Newton's second law, Fnet=maFnet=ma,
so a=Fnet/m=24.64/5.0=4.93 m/s2a=Fnet/m=24.64/5.0=4.93m/s2.

Q5: State Newton's First Law of Motion.
ANSWER ✓ An object at rest stays at rest and an object in motion stays in motion with
the same speed and in the same direction unless acted upon by an unbalanced force.

 Explanation: This law is also known as the law of inertia. It defines a reference frame
where the law holds (inertial frame) and introduces the concept of force as an agent that
causes a change in motion (acceleration).

Q6: What is the difference between static and kinetic friction?
ANSWER ✓ Static friction acts on an object at rest relative to the surface, preventing
motion from starting, while kinetic friction acts on an object that is already sliding
relative to the surface.

 Explanation: Static friction force can vary from zero up to a maximum value of μsNμsN.
Kinetic friction force is generally constant and given by μkNμkN. Typically, μs>μkμs>μk.

Q7: A physics student pushes a 15 kg crate across the floor at a constant velocity. If the
coefficient of kinetic friction is 0.4, with what force is the student pushing?
ANSWER ✓ The student is pushing with a force of 58.8 N58.8N.

,  Explanation: Constant velocity means net force is zero, so the pushing force equals the
kinetic friction force.
o Fpush=Ffric=μkNFpush=Ffric=μkN.
o The normal force NN equals the weight mgmg for a horizontal surface.
o Fpush=(0.4)(15 kg)(9.8 m/s2)=0.4∗147=58.8 NFpush
=(0.4)(15kg)(9.8m/s2)=0.4∗147=58.8N.




Unit 2: Mechanics - Work, Energy, and Power
Key Concepts: Work (constant and variable forces), Kinetic Energy, Work-Energy
Theorem, Potential Energy (Gravitational and Elastic), Conservative vs. Non-conservative
Forces, Conservation of Mechanical Energy, Power.

Q8: How much work is done by lifting a 2.0 kg book from the floor to a shelf 2.5 m high
at a constant speed?
ANSWER ✓ The work done is 49 J49J.

 Explanation: Lifting at constant speed requires an upward force equal to the weight
(mgmg). Work is force times displacement in the direction of the force.
o W=Fd=mgh=(2.0 kg)(9.8 m/s2)(2.5 m)=49 JW=Fd=mgh=(2.0kg)(9.8m/s2)(2.5m)
=49J.
Q9: A 1000 kg car traveling at 20 m/s brakes to a stop. How much work was done by the
friction force?
ANSWER ✓ The work done by friction was −200,000 J−200,000J (or -200 kJ).

 Explanation: The work-energy theorem states that the net work done on an object
equals its change in kinetic energy.
o Wnet=ΔKE=12mvf2−12mvi2Wnet=ΔKE=21mvf2−21mvi2
o Wnet=0−12(1000)(20)2=−200,000 JWnet=0−21(1000)(20)2=−200,000J.
o The negative sign indicates the force is opposite the direction of motion.

Q10: A spring with a spring constant k=500 N/mk=500N/m is compressed by 0.1 m.
How much elastic potential energy is stored in the spring?
ANSWER ✓ The stored energy is 2.5 J2.5J.