# A #3.0*10^-3# kg truck traveling at 20.0 m/s in a test laboratory collides into a wall and comes to rest in 0.10 s. What is the magnitude of the average force acting on the truck during the collision?

##### 1 Answer

If the given mass value is correct, the magnitude of the average force is

#### Explanation:

Impulsive force (

Impulse in terms of the average force:

#J_x=vecF_(avg)*Deltat#

Because we know that

#Deltap=vecF_(avg)Deltat#

Recall that momentum is given by the product of an object's mass and velocity. Therefore, the change in momentum is given by:

#Deltavecp=mv_f-mv_i#

Which is equivalent to

#=>m(vecv_f-vecv_i)=vecF_(avg)Deltat#

Solving for the average force:

#=>vecF_(avg)=(m(vecv_f-vecv_i))/(Deltat)#

We can now plug in the known values:

#=>vecF_(avg)=(3.0*10^-3kg*(0-20.0m/s))/(0.10s)#

#=>vecF_(avg)=-0.6N# Or

#0.6N# opposite the direction that the truck was traveling.

This is a small force, but the mass value of the truck was quite small (

Note that the average force will be less than the maximum force,