Question

Девочка массой 30 кг стоит на льду и бросает своей подруге мяч массой 0.5 кг. Какое максимальное ускорение она может сообщить мячу, оставаясь в покое?Коэффициент трения о лёд равен 0,015.(Указание:Для того что бы девочка оставалась на месте после броска, сила возникающая за счёт отдачи, должна быть меньше силы трения.) Помогите пожалуйста!!! С подробным объяснением

Answer 1

M1=30 кг
m2=0,5 кг
k=0,015

a(max)=?

a=F/m2. Всегда ускорение тела - сила, действующая на тело, деленная на массу тела.
F< или =Fтр. По условию, сила отдачи должна быть меньше или равна силе трения.
Fтр=kN=km1g=0,015 * 30 кг * 10 Н/кг=4,5 Н. Сила трения равна коэффициенту трения, умноженному на силу реакции опоры, которая на горизонтально ровных поверхностях равна весу тела.
Теперь можно найти максимальное ускорение, посчитав силу отдачи равной силе трения.
a=4,5 Н / 0,5 кг=9 м/с2.

Answer 2

Calculation of Maximum Acceleration

To determine the maximum acceleration that the girl can impart to the ball while remaining at rest, we need to consider the forces acting on the system.

The forces acting on the girl-ball system are: 1. The force exerted by the girl on the ball when she throws it. 2. The force of friction between the girl's feet and the ice.

To calculate the maximum acceleration, we need to find the maximum force that the girl can exert on the ball without causing herself to move. This force should be less than the force of friction between the girl's feet and the ice.

Let's calculate the maximum force that the girl can exert on the ball:

1. The force exerted by the girl on the ball is equal to the change in momentum of the ball. Since the girl is at rest before and after throwing the ball, the change in momentum is given by:

Δp = m * v

Where: - Δp is the change in momentum - m is the mass of the ball - v is the velocity of the ball

Since the girl imparts the maximum force to the ball, the ball will have the maximum velocity just after being thrown. Therefore, we can rewrite the equation as:

F = m * v / t

Where: - F is the force exerted by the girl on the ball - t is the time taken to throw the ball

2. The force of friction between the girl's feet and the ice is given by:

F_friction = μ * N

Where: - F_friction is the force of friction - μ is the coefficient of friction between the girl's feet and the ice - N is the normal force (equal to the weight of the girl)

Now, let's calculate the maximum acceleration:

1. The maximum force that the girl can exert on the ball without causing herself to move is equal to the force of friction:

F <= F_friction

Substituting the values, we get:

m * v / t <= μ * N

2. The maximum acceleration that the girl can impart to the ball is given by:

a = F / m

Substituting the value of F from the previous equation, we get:

a <= μ * N / m

Now, let's substitute the given values and calculate the maximum acceleration.

Given: - Mass of the girl (m_girl) = 30 kg - Mass of the ball (m_ball) = 0.5 kg - Coefficient of friction (μ) = 0.015

We need to calculate the maximum acceleration (a).

Calculation:

1. Calculate the force of friction (F_friction): - The normal force (N) is equal to the weight of the girl, which is given by: N = m_girl * g Where g is the acceleration due to gravity (approximately 9.8 m/s^2). - Substituting the given values, we get: N = 30 kg * 9.8 m/s^2 - Calculate the force of friction: F_friction = μ * N Substituting the given value of μ and the calculated value of N, we get: F_friction = 0.015 * (30 kg * 9.8 m/s^2)

2. Calculate the maximum acceleration (a): - The maximum acceleration is given by: a <= F_friction / m_ball Substituting the calculated value of F_friction and the given value of