A box rests on top of a flat bed truck. The box has a mass of m = 15 kg. The coefficient of static friction between the box and truck is μs = 0.76 and the coefficient of kinetic friction between the box and truck is μk = 0.61.
1) The truck accelerates from rest to vf = 17.0 m/s in t = 13.0 s (which is slow enough that the box will not slide). What is the acceleration of the box?
2) What is the frictional force the truck exerts on the box?
3) What is the maximum acceleration the truck can have before the box begins to slide?
4) Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box?
5) With the box still on the truck, the truck attains its maximum velocity. As the truck comes to a stop at the next stop light, what is the magnitude of the maximum deceleration the truck can have without the box sliding?

Explanation:

1.

vf = vi + a(t)

vi = 0 m/s

a = 17/13

= 1.3 m/s^2

2.

F = M * a

= 15 * 1.3

= 19.62 N

3.

Normal force, Fn = m * g

= 15 * 9.8 m/s²

= 147 N.

Then find the maximum force of friction, knowing that μs = 0.76

Ff = Fn × μs

= 147 * 0.76

= 111.83 N

Maximum acceleration,

Ff = m × a

111.83 = 15 * a

a = 7.4556 m/s²

4.

In order to find the acceleration for the box, you need to know the net force of the box moving in the x direction and the frictional force, and you will end up with the Force of the vehicle minus the Frictional force (Ff) between the box and the vehicle, resulting your net force in the x direction.

F = m*a and Ff = μ * N

m*a = μk * N

m*a = μk * m * g

a = μk * g

a = 0.61 * 9.81

5.98 m/s^2 for the acceleration of the box on top of the vehicle.

5.

Assuming the box has re settled and is no longer sliding when braking begins and the surface remains horizontal, the maximum negative acceleration will again be

a = -7.4556 m/s².

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

seeing the graph would