52.(52G) A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed of the crate?

Answer 1

find the final speed at the end of the first 15 m...

vf^2=2ad where a=force/mass
solve for Vf. Then use that as Vi here

vf^2=vi^2+2ad where a= (F-mu*110g)/mass
and d is again 15m
solve for vf.

Answer 2

7. ( 11E ) a) Find the momentum of an automobile of mass 2630 kg traveling 21.0 ms .b) Find the velocity ( in km/h ) of a light auto of mass 1170 kg so that it has the same momentum as the auto in part ( a ).

Answer 3

From a to b
Wg= Fx cos 90 = 0
Wn= Fx cos 90 = 0
Wnet=Efk - Eik. Which is = to 1/2mVf^2 -1/2 mVi^2.
Fxcos(0)= Efk. Cos of 0 =1. Vi =0 so use Efk
Fx=1/2mVf^2
2Fx/m = Vf^2
Vf^2= square root of2Fx/m
Vf= 9.77m/s
X = distance, m= mass, F= force, V= velocity
Efk = final kinetic energy and is 1/2 mv^2
Eik= initial kinetic energy
Fn= normal force, Fg, force gravity
B to C
Fg=0
Fn= 0
Wnet = Efk - Eik
Fxcos(0) + Ffrxcos(180). Cos of zero =1 cos of 180 =-1
Fr= 0.30(9.8)(110)(15)(-1). =-4851
Fx - 4851 = Efk - Eik
350 (15)-4851 = Efk - Eik
399= 1/2mVc^2 - 1/2 m Vb^2
399 = 1/2(110)Vc^2 -1/2(110)(9.8)^2
399 = 1/2(110)Vc^2 -5282
399-5282 = 55 Vc^2
399 + 5282/55 = Vc^2
Vc = square root of 399 + 5282/55
Vc=10.2m/s