AP Physics 1:4.2 Work and Mechanical Energy- Exam Style questions with Answer- MCQ

Question

An object of mass m starts at a height of \(H_1\) with a speed of \(v_1\) . A few minutes later, it is at a height of \(H_2\) and a speed \(v_2\) . Which of the following expressions best represents the work done to the mass by nongravitational forces to the object during this time?
        (A)\(mg(H_{2}-H_{1})+ \frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\)
        (B) \(mg(H_{2}-H_{1})-\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\)
        (C) \(\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\)
        (D) \(\frac{1}{2}m(v_{1}^{2}-v_{2}^{2})\)

▶️Answer/Explanation

Ans:  (A) 

Mechanical energy \(= mgh+\frac{1}{2}mv^{2}\)

Energy “lost” or “gained”\(=mgH_{2}+\frac{1}{2}mv_{2}^{2}-(mgH_{1}+\frac{1}{2}mv_{1}^{2})\)

Work by forces other than gravity = change in energy

QUESTIONS (A) AND (B) ARE BASED ON THE FOLLOWING INFORMATION:
A variable force acts on a 2-kilogram mass according to the graph below.

Question(A)

 How much work was done while displacing the mass 10 meters?
       (A) 40 J
       (B) 38 J
       (C) 32 J
       (D) 30 J

▶️Answer/Explanation

Ans: (C) Work is the area under the curve:
         (4 N)(6 m) + (2 N)(4 m) = 24 J + 8 J = 32 J

Question(B)

 What was the average force supplied to the mass for the entire 10-meter displacement?
       (A) 3.2 N
       (B) 1.2 N
       (C) 4.4 N
       (D) 4 N

▶️Answer/Explanation

Ans: (A ) work = Faverage · displacement
                   32 J = ( Faverage )(10 m)
            Faverage = 3.2 N

Question

What is the work done by a horizontal spring (spring constant k ) expanding from a compression distance x to an extension distance x to an attached mass?
        (A) 2 kx2
        (B) ½ kx2
        (C) kx2
        (D) 0

▶️Answer/Explanation

Ans: (D) \(\frac{1}{2}kx^{2}\) of work is delivered to the mass while uncompressing, followed by − \(\frac{1}{2}kx^{2}\) done while the mass extends the spring outward, totaling to 0 net work for the entire expansion. Alternatively, think about the velocity being zero and the beginning and end of that  single oscillation. No change in kinetic energy occurs; therefore, no net work is done.

Question

A block ( m = 1.5 kg) is pushed along a frictionless surface for a distance of 2.5 meters, as shown above. How much work has been done if a force of 10 newtons makes an angle of 60 degrees with the horizontal?
(A) Zero
(B) 12.5 J
(C) 21.6 J
(D) 25 J

▶️Answer/Explanation

Ans:(B) Work = fd cos 60° = (10 N)(2.5)(0.5) = 12.5 J

Question

Which of the following forces does not do work in its given situation?
(A) Normal force as a person goes up in an elevator
(B) Frictional force as a box slides down a ramp
(C) Centripetal force as a car drives around a circular track
(D) Electrical force as a positive charge moves toward a negative charge

▶️Answer/Explanation

Ans:C
The basic formula for work is W = Fdcosθ, where θ is the angle between the force and the direction of motion. Centripetal force, by definition, is always perpendicular to motion. Therefore, the θ will always be 90, and cos90 = 0, which means this force cannot do work

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