The pickup accelerates towards right. The box is sticky to the pickup, thus its acceleration is the same, towards right. Its inertia (force) is oposing the acceleration, thus it is towards left. For the box not to move, it is necessary that the truck acts on it with a force towards right. (The two forces of the truck on box and the box inertia (force) equilibrate themselves).
Answer:
a = 1,008 10⁻³ m / s²
Explanation:
For this exercise, let's use the kinematic relations of accelerated motion
v² = v₀² - 2 a x
The negative sign is because the acceleration is opposite to the speed, the final speed is zero
0 = v₀² - 2 a x
a = v₀² / 2x
Let's reduce the magnitudes to the SI system
x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m
Let's calculate
a = 2.2²/2 2.4 10⁻³
a = 1,008 10⁻³ m / s²
Force = 3200 N
Work done = 640, 000 Nm
Explanation:
We begin by calculating the deceleration of the truck, using the velocity and distance;
a = (v² – u²)/2s
whereby;
a = acceleration
v = initial velocity
u = initial velocity
s = distance
We begin by changing the speed from km/h into m/s;
54km/hr = 15m/s
Then acceleration;
a = (0² – 15²) / 2 * 200
a = -225 / 400
a = - 0.5625 m/ s²
To calculate force;
F = ma
Whereby;
F = force
M = mass (in kgs)
a = acceleration
F = 1800 / 0.5625
F = 3200 N
Work done = Force * displacement
Work done = 3200 * 200
= 640, 000 Nm
Learn More:
For more on force and work done check out;
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Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
Answer:
Separation increases at all times that rock X falls because it falls with a greater speed
Explanation:
For both rocks, let initial velocity ∪=0
To find the displacement at any given time interval of Δt then
S= ∪Δt +0.5gΔt²
Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y
