Question

A 100 kg box is pulled 10 m across a
frictionless horizontal surface by a 50 N force.
The change in the P.E. of the box is
(a) OJ
(b) 2 J
(c) 20 J
(d) 50 J​

Answer 1

L=2m,

L=2m,d=3mm,A=

L=2m,d=3mm,A= 4

L=2m,d=3mm,A= 49π

L=2m,d=3mm,A= 49π

L=2m,d=3mm,A= 49π ×10

L=2m,d=3mm,A= 49π ×10 −6

L=2m,d=3mm,A= 49π ×10 −6 m

L=2m,d=3mm,A= 49π ×10 −6 m 2

L=2m,d=3mm,A= 49π ×10 −6 m 2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .

L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .

Answer 2

Answer:

Zero J

P E= mgh

Here, h =0

soP.E= 0