Question

Брусок массой 0,8 кг скользит по горизонтальной поверхности . какова сила трения скольжения действующая на брусок если коэффициент трения поверхности бруска равен 0,6

Answer 1

Сила трения (Fтр) равна произведению силы нормального давления, которое прижимает тело к опоре (F), и коэффициента трения (μ);
Fтр = F*μ;
F = m*g;
F = 0.8кг * 10м/c^2 = 8Н:
Fтр = 8Н * 0.6 = 4.8Н.

Answer 2

Calculation of the Sliding Friction Force

To calculate the sliding friction force acting on the block, we can use the formula:

F = μN

Where: - F is the sliding friction force - μ is the coefficient of friction - N is the normal force

In this case, the block has a mass of 0.8 kg. The normal force can be calculated as the product of the mass and the acceleration due to gravity:

N = mg

Substituting the given values, we have: - m = 0.8 kg - g = 9.8 m/s^2

Therefore: - N = 0.8 kg * 9.8 m/s^2 = 7.84 N The coefficient of friction for the surface of the block is given as 0.6.

Substituting the values into the formula, we have: - μ = 0.6 - N = 7.84 N

Therefore: - F = 0.6 * 7.84 N = 4.704 N

So, the sliding friction force acting on the block is approximately 4.704 N.