what is the rate of acceleration of a 2,000-kg truck if a force of 4,200n is used to make it start moving forward
1 answer:
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Answer:
F = 20.4 i ^
Explanation:
This exercise can be solved using the ratio of momentum and amount of movement.
I = F t = Dp
Since force and amount of movement are vector quantities, each axis must be worked separately.
X axis
Let's look for speed
cos 45 = vₓ / v
vₓ = v cos 45
vₓ = 8 cos 45
vₓ = 5,657 m / s
We write the moment
Before the crash p₀ = m vₓ
After the shock = -m vₓ
The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ
The impulse on the x axis Fₓ t = Δp
Fₓ = 2 m vₓ / t
Fx = 2 0.450 5.657 / 0.250
Fx = 20.4 N
We perform the same calculation on the y axis
sin 45 = vy / v
vy = v sin 45
vy = 8 sin 45
vy = 5,657 m / s
We calculate the initial momentum po = m
Final moment = m
Variations moment Δp = m - m
= 0
Force in the Y-axis = 0
Therefore the total force is
F = fx i ^ + Fyj ^
F = Fx i ^
F = 20.4 i ^
Answer:
25.59 m/s²
Explanation:
Using the formula for the force of static friction:
--- (1)
where;
static friction force
coefficient of static friction
N = normal force
Also, recall that:
F = mass × acceleration
Similarly, N = mg
here, due to min. acceleration of the car;
From equation (1)
However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.
Thus,
where;
and g = 9.8 m/s²