Question

Для равномерного передвижения порожнего грузовикана расстояние 100 м сила тяги его двигателя выполнила работу в 1,6 МДж а после загрузкигрузовика- 2 МДж. во сколько раз выросла сила трения во втором случае

Answer 1

Для равномерного передвижения порожнего грузовикана расстояние 100 м сила тяги его двигателя выполнила работу в 1,6 МДж а после загрузкигрузовика- 2 МДж. во сколько раз выросла сила трения во втором случае

А1 = F1*S
A2=F2*S
А2/А1=F2/F1
F2/F1=2/1,6=1,25
Сила трения увеличилась в 1,25 раз

Answer 2

Calculation of the Force of Friction

To determine the increase in the force of friction when a truck is loaded, we need to compare the work done by the engine in both cases.

In the first case, when the truck is empty, the engine performs work equal to 1.6 MJ to move the truck a distance of 100 m. In the second case, after the truck is loaded, the engine performs work equal to 2 MJ to move the loaded truck the same distance.

We can use the work-energy principle to calculate the force of friction in both cases. According to the work-energy principle, the work done by a force is equal to the change in kinetic energy. In this case, the work done by the engine is equal to the change in kinetic energy of the truck.

Calculation:

In the first case, when the truck is empty: - Work done by the engine = 1.6 MJ - Distance traveled by the truck = 100 m

In the second case, after the truck is loaded: - Work done by the engine = 2 MJ - Distance traveled by the truck = 100 m

Since the distance traveled is the same in both cases, we can equate the work done by the engine to the change in kinetic energy of the truck:

1.6 MJ = Change in kinetic energy (Case 1) 2 MJ = Change in kinetic energy (Case 2)

To find the increase in the force of friction, we need to compare the change in kinetic energy in both cases. Since the work done by the engine is directly related to the force of friction, we can say that the increase in the force of friction is directly proportional to the increase in the work done by the engine.

Therefore, the force of friction in the second case is 2/1.6 = 1.25 times greater than the force of friction in the first case.

Answer:

The force of friction in the second case, after the truck is loaded, is 1.25 times greater than the force of friction in the first case, when the truck is empty.