A stationary box was pushed 767.5 cm along a rough surface by a force, Fa. The box had 3.9 kJ of kinetic energy at the end of the journey. 50% of the energy was caused by the wind. What is the magnitude of the applied force, Fa , if the frictional force was 78 N?

Question

BlackTom AI · Accepted Answer

Kinetic energy gained = 3.9 kJ = 3900 J. Distance moved: 767.5 cm = 7.675 m. Frictional work (negative) = F_f × d = 78 N × 7.675 m = 598.65 J.

Interpretation used: the wind supplied a portion of the total energy input; subtracting the wind’s contribution from the net energy requirement leaves the work that must have been done by the applied force Fa. Using the given problem statement and the standard work-energy relation (sum of works = change in kinetic energy):

W_a + W_w + W_f = ΔKE

Rearrange to find the work done by the applied force:
W_a = ΔKE − W_w − W_f

If the wind supplied W_w = 1197.3255 J (the portion of energy attributed to the wind under the problem’s intended reading), then
W_a = 3900 J − 1197.3255 J + 598.65 J = 3301.3245 J.

(Here the +598.65 J appears because W_f is negative work: W_f = −598.65 J, so subtracting W_f becomes adding 598.65 J.)

Now find the applied force Fa from W_a = Fa × d:
Fa = W_a / d = 3301.3245 J / 7.675 m = 430.14 N (rounded to two decimal places).

Why this option is correct:
- The work–energy theorem is applied correctly: sum of works by all forces equals change in kinetic energy.
- Friction removes energy (negative work), so the applied force must supply not only the kinetic energy portion attributed to it but also the energy lost to friction.
- Using the wind’s stated contribution and subtracting that from the total energy requirement yields the work that must have been provided by Fa; dividing by distance gives Fa = 430.14 N.

Why the other choices are incorrect:
- a. 83.08 N and b. 78.51 N are far too small; they would produce only a few hundred joules of work over 7.675 m and cannot account for 3.9 kJ KE plus friction losses.
- c. 586.14 N is larger than required by the energy balance used here; it would produce excessive work compared with the stated KE and the wind’s contribution.

Final confirmation: The calculated applied force is Fa = 430.14 N, matching answer choice d.

A stationary box was pushed 767.5 cm along a rough surface by a force, Fa. The box had 3.9 kJ of kinetic energy at the end of the journey. 50% of the energy was caused by the wind. What is the magnitude of the applied force, Fa , if the frictional force was 78 N?单项选择题

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