Question

KE* = /2my? (m) -247 PROBLEM 4-8 points (2 points each) Fx (N) A wheeled cart is free to move along the +47 x-axis. The cart has a mass of 4.00 kg and an initial velocity of 4.00 m/s in the +x- direction when it is at x = Om. If the cart + + + + + is between x -1 m and x = +2 m, the net force is 1.00 N in the positive x- direction. If the cart is between x +2 m and x = +7 m, the net force is 4.00 N in the negative x-direction. If the cart is between x = +7 m and x +8 m, the net force is 2.00 N in the positive x-direction. The net force is zero at all other locations. position (m) Kinetic energy 0) (a) Complete the table, to show the cart's 10 kinetic energy at 1-meter intervals. 3 (b) What is the cart's minimum kinetic energy? At what location is the cart when its kinetic energy is this minimum value? 5 (c) What is the network done on the cart over the 8-meter distance? +4.5 m. What initial velocity would (d) Let's say that you wanted the cart to reverse direction at x the cart have to have at x 0 m for this to happen?

Answer 1

from Benton's equation of motien, w? = U²+ zas for it to a em f = IN 2) a = € I 0.25 m/s2 4 for r2 #zm to r = +7m, fnet = -4 N. -- =-mis? for i= + 7 to not 8 m, fnet = 2.0N -932_ 0.5 m/s2 a u= 4 m/s at 20 I so KE = + mv² = I x 4 x 16 KE. z 32 I (1:0)

и = (4)2 + 2ҳо 25x 1 и“ - 16.5 - Кs - 1 my 16.5 х 4 . 33 J (и - Ім) иг (4)*+ 2 xo. 25х2 и? - 17 KS- 2 m/ ə k. E 2 34 т. ( 2 т л о с . - м . r C # 2 to 7 m а : -1 м/с- “. 17 + 2х(-1)xІ - 2 15 с. Е. = 30 т. к. 3 mЯ у*. 11 + 2 (-1) х2 у? 2 13 k. E. a 2 ( 1 кат. 4 ту уг 17 + 2(-1)x3 = 11 Е. Е. - 22T (из 5 my у*. 1+ + 2(-1)хи : 3 к.Е. 19. мг 6 т * : 11 - 10 27 te. E. . 14 л (157mУ оY M2 +7 to 312-+8 м , а... 0.5 м 2 = 8 2 7 +22х0.5x| < . . . . 14 л/

Position (m) K: . 32 33 34 30 26 22 18 14. 16 (K Elmin = 14 J ? r = 7 m / c) Whet a (f) displacement Wnet - (1 2X2 + (-4)x5 +(2) (1) Wnet = -16 J Tave sign shums that worlo in done en the cary 2 for o to a m u² = U² + 2x0.25x2 1² = U²+1 T U is initial velocity at Holy for 2 to 4.5 m o. ho g. 120 at 4.5 my 02 (0+1) + 2x(+1)x 215 0 = U²+1 - 5 0224 702 2 m/s

Answer 2

(a) At x=0m the velocity is 4m/s and the KE:

$KE=1/2mv_{i}^{2}$

$KE(0m)=1/2\times 4kg\times (4m/s)^{2}=32J$

For x=1m we must use the work-energy theorem:

$W_{F_{net}}=\Delta KE$

$Fd=KE(1m)-KE(0m)$

$1N\times 1m=KE(1m)-32J$

$KE(1m)=33J$

Following the same reasoning:

$KE(2m)=34J$

Now between x=2m and x=3m the force changes:

$-4N\times 1m=KE(3m)-34J$

$KE(3m)=30J$

Following the same reasoning:

$KE(4m)=26J$

$KE(5m)=22J$

$KE(6m)=18J$

$KE(7m)=14J$

Now between x=7m and x=8m the force changes again:

$2N\times 1m=KE(8m)-14J$

$KE(8m)=16J$

(b) The minimum is KE=14J at x=7m

(c) The net work done on the cart over an 8 m distance is given by:

$W_{F_{net}}=KE(8m)-KE(0m)$

$W_{F_{net}}= 16J-32J=-16J$

(d) The net work done from x=0m to x=4.5m is:

$1N\times 2m&plus;(-4N)\times 2.5m=-8J$

Because of the WE theorem:

$-8J=KE(4.5m)-KE(0m)$

Since we want the cart to stop at 4.5m, it's KE must be zero at that point, then:

$8J=KE(0m)=1/2mv_{i}^{2}$

$v_{i}=\sqrt{\frac{16J}{4kg}}=2m/s$