A pickup truck is carrying a 10.0 kgkg toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts moving. The coefficients of static and kinetic friction between the toolbox and the truck bed are 0.30 and 0.13, respectively. Assume that the truck bed is horizontal. What is the shortest time the truck could take to accelerate from rest to 10.2 m/sm/s without causing the toolbox to slide? A pickup truck is carrying a 10.0 toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts moving. The coefficients of static and kinetic friction between the toolbox and the truck bed are 0.30 and 0.13, respectively. Assume that the truck bed is horizontal. What is the shortest time the truck could take to accelerate from rest to 10.2 without causing the toolbox to slide? 3.47 ss 8.00 ss 2.60 ss 2.25 ss

Shortest time = 3.47 seconds

Explanation:

We know that Force of gravity of an object is given by;

F_g = mg

Also, Force due to kinetic friction is given as;

F_friction = μ_k•F_n

Force due to static friction is given as;

F_static = μ_s•F_n

Where ;

μ_k = coefficient of kinetic friction

μ_s = coefficient of static friction

F_n = Normal Force

Let's solve for the static crate.

From the free body diagram i attached, the resultant force in the vertical y direction will be;

N - mg = 0

Thus, N = mg

WherN is the

From earlier, we know that;

F_static = μ_s•F_n

Thus,

F_static = μ_s•mg

Now, in the horizontal x direction,

F_static = ma

Thus,

μ_s•mg = ma

m will cancel out, and;

a = μ_s•g

a = 0.3 x 9.81 = 2.943 m/s²

Not, to get the final the final speed of the tool box, we'll use;

V_f = V_i + a•Δt

Where;

Δt is shortest time.

V_f = final velocity

V_i = initial velocity

Thus, making Δt the subject, we have;

Δt = (V_f - V_i)/a = (10.2 - 0)/2.943 = 3.47 s

its a 5.5 n

glad that