📝 Work done by constant force calculus (32 MCQs)

📖 Explanation: The unit of work is the joule (J), which is equivalent to a newton-meter (N·m). Since a newton is kg·m/s², multiplying by meters gives kg·m²/s². Options A and D represent other physical quantities (pressure and power, respectively). Option B is the definition of a newton, not work.

📖 Explanation: Work done by a constant force is path-independent and depends only on the force and displacement in the direction of the force, not on the presence of other forces like friction. The student applies the same force (50 N) over the same displacement (4 m), so the work done is W = Fd = 50 × 4 = 200 J. Friction does negative work, but that doesn't change the work done by the student.

📖 Explanation: Since the load moves at a constant speed, the tension force in the cable equals the weight of the load: T = mg = 500 × 10 = 5000 N. The displacement is upward, same as the force, so work is positive: W = Td = 5000 × 10 = 50,000 J. The gravitational force does -50,000 J, but the question asks specifically for work done by tension.

📖 Explanation: Work is defined as W = Fd cos θ, depending on displacement (d), not time. Impulse (J = FΔt) depends on time. Two objects subjected to the same force for the same time will have the same impulse, but if they have different masses or initial velocities, their displacements (and thus work) can be very different. This is a common misconception.

📖 Explanation: For a force that is constant (horizontal line on a force vs. position graph), the work done is the area under the curve, which is a rectangle of width 5 m and height 10 N, giving 50 J. Option A is incorrect because the graph gives enough information. Option C confuses work with the concept of a spring constant (slope of force vs. displacement for a spring).

📖 Explanation: Work requires displacement. Since the book is stationary (displacement = 0 m), the work done by any force (including the person's upward force) on the book is zero, even if a force is being exerted. This is a classic trick question to test the definition of work, distinguishing it from effort or force.

📖 Explanation: Work is W = Fd cos θ. Here, F = 20 N, d = 5 m, and θ = 60°. cos 60° = 0.5. Thus, W = 20 × 5 × 0.5 = 50 J. A common error is to use sin 60° instead of cos 60°, or to ignore the angle entirely and multiply 20 × 5 = 100 J.

📖 Explanation: The formula W = Fd only applies to constant forces. For a variable force (including one that varies linearly), the work is the area under the force vs. position curve, given by the integral W = ∫ F(x) dx. Option D is the kinetic energy formula. This tests the Easy of when the simple formula is applicable and when calculus is needed.

📖 Explanation: Work done by applied force: 10 × 3 = 30 J. Friction force = μmg = 0.2 × 2 × 10 = 4 N. Work done by friction = -4 × 3 = -12 J. Normal and gravity do zero work. Total work = 30 - 12 = 18 J. This requires applying the work formula to multiple forces and summing, a higher-order skill.

📖 Explanation: By the work-energy theorem, the net work done on an object equals the change in its kinetic energy (½mv²). If velocity is constant, kinetic energy is constant, so the change is zero, and net work must be zero. This is true even if individual forces do work, because their sum is zero. Option C is a subtle misconception, as balanced forces can result in zero net work while individual forces may be non-zero.

📖 Explanation: The gravitational force is F = mg = 50 N downward. The displacement is 4 m upward. The angle between force and displacement is 180°, so W = Fd cos 180° = 50 × 4 × (-1) = -200 J. Work done by gravity is negative when lifting. It is independent of the speed (as long as it's constant).

📖 Explanation: Using the work-energy theorem: W_net = ΔK. Work = Fd = F(50). ΔK = ½ × 1000 × (20)² = 200,000 J. So, 50F = 200,000 => F = 4000 N. This is an Medium of the work-energy theorem to find a force, requiring more steps than direct use of W = Fd.

📖 Explanation: Work is defined as force times displacement in the direction of the force. If the displacement is zero (the crate doesn't move), the work done is zero, regardless of how much force is applied. The person may be exerting a lot of effort, but physically, no work is done on the crate. This emphasizes the crucial distinction between effort and work.

📖 Explanation: This is a 'single-step' but conceptual problem. The net force is zero (5 N east + 5 N west = 0). Therefore, the net work done is also zero. Alternatively, work by east force = 5 × 10 = 50 J, work by west force = 5 × 10 × cos 180° = -50 J. Total work = 0 J.

📖 Explanation: This addresses a very common misconception about simple machines. While a ramp reduces the required force, the distance increases proportionally. In an ideal scenario (no friction), the work done (W = Fd) remains the same, equal to mgh. Option C is true for work against gravity, but option B provides the direct explanation in terms of force and distance, directly addressing the student's reasoning.

📖 Explanation: We apply W = Fd cos θ. The displacement d = 7 - 2 = 5 m. F = 6 N, θ = 30°. cos 30° = √3/2 ≈ 0.866. W = 6 × 5 × 0.866 = 25.98 J. This question adds a displacement that isn't from zero, requiring the student to correctly calculate the displacement first, which is a key step.

📖 Explanation: Work is the area under the force vs. displacement curve. The rectangle area = 4 × 5 = 20 J. The triangle area = ½ × 4 × 5 = 10 J. Total area = 30 J. This tests graph interpretation by combining geometric shapes, beyond a simple constant force graph.

📖 Explanation: Use the work-energy theorem: W_net = ΔK. The net work is Fd (since only one force). d = 12 m. ΔK = ½ × 3 × (8² - 4²) = 1.5 × (64 - 16) = 1.5 × 48 = 72 J. Then, 12F = 72 => F = 6 N. This is a two-step problem requiring conversion of velocity to kinetic energy and then to force.

📖 Explanation: Work done by the force equals the change in kinetic energy. Initial K = ½mv². Final K = ½m(2v)² = 2mv². Change in K = 2mv² - ½mv² = 3/2 mv². This does not require knowing the force or distance. It tests the conceptual link between work and change in kinetic energy, not just the formula W=Fd.

📖 Explanation: W = Fd cos θ = 8 × 5 × cos 120°. cos 120° = -0.5. W = 40 × (-0.5) = -20 J. This tests recall of the cosine of a non-standard angle (120°) and understanding that work can be negative. The negative sign indicates the force opposes motion.

📖 Explanation: The student's statement is incomplete and potentially incorrect. Work is done only when the force has a component in the direction of the displacement. If the student carries the backpack up the stairs, the force is upward (to support the weight), and the displacement is upward, so positive work is done by the student on the backpack. If the student walked horizontally, the force and displacement would be perpendicular, and the work would be zero. The student needs to specify the relative directions.

📖 Explanation: If the block moves with constant velocity, the net force is zero (F - f = 0). Therefore, the magnitude of friction is equal to the applied force. Since the displacements are the same and opposite in direction, the work done by friction is -Fd, while the work done by F is +Fd. This question reinforces the link between constant velocity and zero net work.

📖 Explanation: Work is area under the F-x graph. Area of rectangle (0-2 m): 2 × 4 = 8 J. Area of triangle (2-6 m): base = 4 m, height = 4 N. Area = ½ × 4 × 4 = 8 J. Total work = 8 + 8 = 16 J. This requires calculating areas of both a rectangle and a triangle and then summing.

📖 Explanation: W = Fd cos θ = 15 × 2.5 × cos 40°. cos 40° ≈ 0.766. W = 37.5 × 0.766 ≈ 28.7 J. This is a straightforward Medium but requires knowledge of the cosine of 40° and proper use of the angle between force and displacement.

📖 Explanation: Work depends on the angle between force and displacement: W = Fd cos θ. Here, Fd = 50 J. Since θ can be any angle from 0° to 180°, cos θ can range from 1 to -1. For work to be positive, θ must be between 0° and 90°. For negative work, between 90° and 180°. Thus, the work can be any value from -50 J to +50 J. Option C is partially correct but misses negative values. Among the given options, C is the best as it captures the idea of dependence on angle.

📖 Explanation: If the forces are identical and the displacements are equal in magnitude but opposite in direction, then the work done is W = Fd cos θ. If the force and displacement are in the same direction for A, work is +Fd. If the force is applied to B in the same direction as its displacement (i.e., the force is also opposite), then work is also +Fd, but the question states 'Box B moves 2 m but in the opposite direction.' This implies the force is applied in a constant direction, say to the right, and Box A moves right, Box B moves left. Then W_A = Fd (positive), W_B = Fd cos 180° = -Fd. So they are equal in magnitude but opposite in sign.

📖 Explanation: Lifting the box: Work against gravity = mgh = 5 × 10 × 1.2 = 60 J. Carrying it horizontally: The force applied by the person is vertical (to counteract gravity), but the displacement is horizontal. The angle between force and displacement is 90°, so the work done is zero. Total work = 60 J. The horizontal speed and force used to move it are irrelevant because work is done only against gravity during the horizontal motion. Work done by the person is positive on the box.

📖 Explanation: Initial work: W = Fd. New force: F_new = 2F. New distance: d_new = d/2. New work: W_new = (2F)(d/2) = Fd = W. This is a direct Medium of the formula W = Fd, requiring a simple proportional reasoning.

📖 Explanation: The normal force acts perpendicular to the displacement. The displacement is horizontal, while the normal force is vertical. Since the angle between normal force and displacement is 90°, the work done by the normal force is W = Nd cos 90° = 0. This is true regardless of the velocity or friction. This question tests the definition of work and the role of perpendicular forces.

📖 Explanation: First, find the maximum height using kinematic equations: v_f² = v_i² + 2ad. 0 = (20)² + 2(-10)h => 0 = 400 - 20h => h = 20 m. Work done by gravity = -mgh = -2 × 10 × 20 = -400 J. This requires a multi-step approach: kinematics to find the displacement, then work formula. The negative sign indicates gravity opposes the upward motion.

📖 Explanation: The graph is a triangle with base 10 and height 10. The area of a triangle is ½ × base × height = ½ × 10 × 10 = 50 J. This is the work done. This reinforces the concept that area under a force-displacement curve represents work.