Question

Брусок массой 5 Кг покоится на шероховатом горизонтальном столе .Коэффициент трения между поверхностью бруска и поверхностью стола равен 0.2 . На этот брусок действует горизонтально направленной силой 2.5 Н . Чему равна по модулю возникающая при этом сила трения ?

Answer 1

Дано   m=5 кг    k=0,2       Fтяг =2,5 Н   Fтр- ?

1) найдем силу трения скольжения
Fтр ск=k*m*g=0,2*5*10=10 Н

2) так как Fтяг < 10 Н ( тело останется в покое)
3) так как тело покоится   Fтяг=Fтр покоя=2,5 Н

Ответ Fтр покоя=2,5 Н

Answer 2

Calculation of the Frictional Force

To calculate the magnitude of the frictional force acting on the block, we can use the formula:

F_friction = μN

Where: - F_friction is the frictional force - μ is the coefficient of friction - N is the normal force

In this case, the coefficient of friction (μ) between the block and the table is given as 0.2, and the normal force (N) is equal to the weight of the block, which can be calculated as the product of the mass (m) and the acceleration due to gravity (g).

Given that the mass of the block is 5 kg, we can calculate the weight as:

Weight (N) = mass (kg) × acceleration due to gravity (m/s²)

Using a standard value of acceleration due to gravity of 9.8 m/s², we can calculate the weight:

Weight (N) = 5 kg × 9.8 m/s² = 49 N.

Now, we can substitute the values into the formula for the frictional force:

F_friction = 0.2 × 49 N = 9.8 N.

Therefore, the magnitude of the frictional force acting on the block is 9.8 N.

Conclusion

The magnitude of the frictional force acting on the block, given a coefficient of friction of 0.2 and a normal force of 49 N, is 9.8 N.