A toy car, with mass of 2 Kg, is pushed with a force of 8 N. If the toy car is in the grass with a coefficient of friction of 0.1 then what is the acceleration? How far does it get pushed in 2 s?
To find the acceleration of the toy car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration (F = m*a).
In this case, the force acting on the toy car is the force applied to it minus the force of friction. The force applied to the car is 8 N. The force of friction can be calculated by multiplying the coefficient of friction (μ) by the normal force (N). In this case, the normal force is equal to the weight of the car, which is the mass (m) multiplied by the acceleration due to gravity (g = 9.8 m/s^2).
The force of friction = μ * N
= μ * (m * g)
= 0.1 * (2 kg * 9.8 m/s^2)
= 0.1 * 19.6 N
= 1.96 N
Therefore, the net force acting on the car is 8 N - 1.96 N = 6.04 N.
Now, we can use Newton's second law to find the acceleration:
F = m * a
6.04 N = 2 kg * a
a = 6.04 N / 2 kg
a = 3.02 m/s^2
So, the acceleration of the toy car is 3.02 m/s^2.
To find how far the car gets pushed in 2 seconds, we can use the equations of linear motion.
The equation to find the distance traveled (d) in terms of initial velocity (u), acceleration (a), and time (t) is:
d = u * t + 0.5 * a * t^2
Since the car starts from rest (u = 0), the equation simplifies to:
d = 0.5 * a * t^2
d = 0.5 * 3.02 m/s^2 * (2 s)^2
d = 0.5 * 3.02 m/s^2 * 4 s^2
d = 0.5 * 3.02 m/s^2 * 16
d = 24.16 m
Therefore, the toy car gets pushed a distance of 24.16 meters in 2 seconds.