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,