A stationary box was pushed 733.9 cm along a rough surface by a force, Fa. The box had 2.2 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 83.7 N?

Question

BlackTom AI · Accepted Answer

Work-energy theorem: the total forward work done by applied forces minus the work done by friction equals the change in kinetic energy. First convert the displacement to metres: s = 733.9 cm = 7.339 m.

Relevant knowns:
- ΔKE = 2.2 kJ = 2200 J
- friction force f = 83.7 N (opposes motion)
- wind supplies 50% of the energy (so the total forward work must supply the KE + overcome friction; the applied force Fa in the question is taken as the total forward force whose work, together with the wind’s contribution, produces the required energy)

Step-by-step calculation:
1. Compute the work done against friction (magnitude): W_fric = f · s = 83.7 N × 7.339 m = 83.7 × 7.339 = 614.2743 J.
2. The total forward work required (work done by all forward agents combined) must cover both the kinetic energy gain and the frictional losses:
   W_total_forward = ΔKE + W_fric = 2200 J + 614.2743 J = 2814.2743 J.
3. If Fa denotes the total applied forward force whose work over distance s equals W_total_forward, then Fa · s = W_total_forward. Solve for Fa:
   Fa = W_total_forward / s = 2814.2743 J / 7.339 m = 383.47 N (rounded to two decimal places).

Why this makes the correct physical sense: the box needed 2814.27 J of forward work in total to end up with 2200 J of kinetic energy after overcoming friction. Dividing that total work by the displacement gives the magnitude of the forward force that produced that work. The statement “50% of the energy was caused by the wind” implies that the wind provided half of the energy input; the total forward work (wind + other applied force) still equals 2814.27 J, which is the value used above to find Fa.

Why the other options are incorrect (briefly):
- a (84.00 N) and b (86.70 N) are far too small to do ~2814 J of work over only 7.339 m.
- c (216.07 N) is also too small; it would produce only about 216.07×7.339 ≈ 1586 J of forward work, insufficient to supply the required 2814 J.

Conclusion: The calculated magnitude of the applied force is 383.47 N, which matches answer d.

A stationary box was pushed 733.9 cm along a rough surface by a force, Fa. The box had 2.2 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 83.7 N?单项选择题

类似问题

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?

A joule (J) is the unit of measure of

Find the work done on a box if it is pushed with a horizontal force of 1.5 N for a distance of 1.2 m. Give your answer to two decimal places.

A frictional force of 5 N west acts on a trolley as it rolls 3.4 m to the east. How much work did the frictional force do on the trolley? Give your answer to the nearest whole number.

A 0.20 kg clockwork toy car, starting from rest, travels 16 cm in 0.21 s with a constant acceleration. How much work is done on it? Give your answer to two significant figures.

How much work input is required to lower (not drop) a 120 pound weight from the top of a 3 foot table to the floor?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单