We can use the kinematic equation

where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance
we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

So the maximum acceleration on the block is

4)As the box slides, it is now subjected to kinetic friction, which is

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2
<span>We can answer this using
the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²
-----> 1</span></span>
ω² = ω₀² + 2αθ
-----> 2
Where:
θ = final angular
displacement = 70.4 rad
θ₀ = initial
angular displacement = 0
ω₀ = initial angular
speed
ω = final angular speed
t = time = 3.80 s
α = angular acceleration
= -5.20 rad/s^2
Substituting the values
into equation 1:<span>
70.4 = 0 + ω₀(3.80)
+ ½(-5.20)(3.80)² </span><span>
ω₀ = (70.4
+ 37.544) / 3.80 </span><span>
ω₀ = 28.406
rad/s </span><span>
Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4
ω = 8.65 rad/s
</span>
In your question where as ask the man lift various loads with the same lever and the distance of the applied force from the fulcrum is 2m and the distance from the fulcorm to load is 0.5m.So the following are the answers to your questions:
a.mechanical advantage of the lever is 1
b..<span>ideal mechanical advantage of the lever is 4</span>
c.4