Question

A 4700 kg truck carrying a 900 kg crate is traveling at 25m/s to the right along a straight, level highway, as shown. The truck driver then applies the brakes, and as it slows down, the truck travels 55 m in the next 3.0 s. The crate does not slide on the back of the truck.

 

(a) Calculate the magnitude of the acceleration of the truck, assuming it is constant. 

(b) On the diagram below, draw and label all the forces acting on the crate during braking.

(c)

i. Calculate the minimum coefficient of friction between the crate and truck that prevents the crate from sliding. 

ii. Indicate whether this friction is static or kinetic. 

____ Static ____Kinetic

Now assume the bed of the truck is friction less, but there is a spring of spring constant 9200N/m attaching the crate to the truck, as shown below. The truck is initially at rest.

(d) If the truck and crate have the same acceleration, calculate the extension of the spring as the truck accelerates from rest to 25 m s in 10 s. 

(e) At some later time, the truck is moving at a constant speed of 25 m/s and the crate is in equilibrium. Indicate whether the extension of the spring is greater than, less than, or the same as in part 

(d) when the truck was accelerating. 

___ Greater ___ Less ___ The same 

Explain your reasoning.

Answer 1

(c) using the diagram above and understanding that the static friction is actually responsible for decelerating the box to match the deceleration of the truck, we apply F_net. 

F_net = ma 

– f_s = – μ_sF_n = ma 

– μ_smg = ma 

– μ_s = a/g

– μ_s = – 4.4/9.8 

μ_s = 0.45

Static friction applied to keep the box at rest relative to the truck bed.

(d) Use the given info to find the acceleration of the truck 

a = ∆v/t = 25/10 = 2.5 m/s²

To keep up with the trucks acceleration, the crate must be accelerated by the spring force, apply F_net

F_net = ma 

F_sp = ma 

k∆x = ma 

(9200)(∆x) = (900)(2.5) 

∆x = 0.24 m

(e) If the truck is moving at a constant speed the net force is zero. Since the only force acting directly on the crate is the spring force, the spring force must also become zero therefore the ∆x would be zero and is LESS than before. Keep in mind the crate will stay on the friction less truck bed because its inertia will keep it moving forward with the truck (remember you don’t necessarily need forces to keep things moving).