A student performs an experiment to determine the distance moved and velocity of a projectile as a function of time. A 1.00-kg object is initially at rest. The student applies a force of 19.6 N through a distance of 0.50 m to propel the object straight upward. It has an initial speed v and reaches a peak height h above the launch point. What is the average acceleration of the object from rest to v?

Question

A student performs an experiment to determine the distance moved and velocity of a projectile as a function of time. A 1.00-kg object is initially at rest. The student applies a force of 19.6 N through a distance of 0.50 m to propel the object straight upward. It has an initial speed v and reaches a peak height h above the launch point. What is the average acceleration of the object from rest to v?

BlackTom AI · Accepted Answer

We start by clearly restating the problem setup: a 1.00-kg object is launched upward with a constant force of 19.6 N applied through a distance of 0.50 m, starting from rest, and we are asked for the average acceleration from rest to the final velocity v.
Option A: 19.6 m/s^2. Since the acceleration is assumed constant during the 0.50 m of propulsion, we can use the kinematic relation v^2 = 2 a s, where s = 0.50 m and v is the final speed after propulsion. To find a, we first relate the applied force and distance to the velocity via the work-energy principle: the work done is W = F s = 19.6 N × 0.50 m = 9.8 J, which yields the final kinetic energy (starting from rest): 1/2 m v^2 = 9.8 J, so v^2 = 19.6 and thus v^2 = 2 a s gives a = v^2 / (2 s) = 19.6 / (2 × 0.50) = 19.6 m/s^2. This matches the maximum acceleration consistent with constant acceleration over the propulsion distance.
Option B: 9.8 m/s^2. This value corresponds to half of the acceleration computed above, which would imply either using half the distance or misapplying a formula. It does not satisfy the kinematic relation v^2 = 2 a s with the given s and the velocity derived from the work done, so it’s not correct.
Option C: 39.2 m/s^2. This equals 2 × 19.6 m/s^2, which would arise if you mistakenly multiplied by an extra factor of 2 or used an incorrect rearrangement of v^2 = 2 a s. It overestimates the acceleration for the given distance and velocity relation, so it misrepresents the motion under the stated conditions.
Option D: 19600 m/s^2. This is vastly larger and corresponds to treating the force or energy in an unrealistic way (for example, converting joules directly to acceleration without considering mass and distance correctly). It ignores the mass, distance over which the force is applied, and the kinematic relation linking acceleration to final velocity, making it physically inconsistent with the scenario.
In summary, Option A is the correct choice because constant acceleration over 0.50 m from rest to velocity v yields a = v^2/(2s), and using the supplied force and distance confirms v^2 = 19.6, hence a = 19.6 m/s^2. The other options arise from common misapplications of energy, distance, or kinematic equations and do not satisfy the given conditions.

PHYS 2010-001 Fall 2025 Block 1 Practice Exam Physics 2010 Fall 2025

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