Question

that varies with position as shown in the following figure. A 5.4 kg box is subject to a force F. (N) NO 0 2 4 6 8 10 12 x(m) 16 14 a. Find the work done by the force on the particle as it moves from x = 0 to x = 2.00 m. b. Find the work done by the force on the particle as it moves from x 5.00 m to x 6.00 m. C. Find the work done by the force on the particle as it moves from x 120 m to x = 15.0 m. d. What is the total work done by the force over the distance x = 0 to x = 15.0 m? e. Determine the speed of the box when it is at a position of 15 meters. 1. If the movement took a total of 12 seconds, determine the power exerted,

Answer 1

W = ∫ F.dx = Area of F -x curve so

A)

Area between x = 0 to 2

W = 1/2*base *height = 1/2*2*1 = 1 J answer

B)

W = area of rectangle = (6-5) *3 = 3 J ans

C)  
W = 1/2*( 15-12) *1.5 = 2.25 J ans

D)

w = 1/2*5*3+ ( 10-5) *3 + 1/2* ( 15-10) *3 = 30 J

E)

W = change in KE  

30 = 1/2*m*v^2 = 1/2*5.4*v^2

v≈3.33333 m/s ans

F)

P = W/t = 30/12 =2.5 Watt