Conservation of Engery Physics problem?

Question

A man swings on a 30-m rope initially inclined at an angle of 37 degress with the vertical. What is is speed at the bottom of the swing? a) if starts at rest? b) if he starts with a initial speed of 4 m/s ?

Answer 1

the man drops a vertical displacement of
30*(1-cos(37) m
call this h
his kinetic energy is
.5*m*v^2
which is equal to starting kinetic plus loss of potential
.5*m*vi^2+m*g*h
the equation is
v^2=vi^2+2*g*h

in a, vi=0 and b vi=4, h is the same for both

j

Answer 2

The height h through which he moves downwards is

h = 30 (1 - cos 37) m = 6.0409... m. ......(A)

The equation governing both motions is

v^2 = u^2 + 2gh, where u is the initial spped and v is the final speed.

(a) u = 0; then v^2 = 2*9.81*6.0409... m^2/s^2,

= 118.523... m^2/s^2, so that

v = 10.887... m/s.

(b) u = 4 m/s; then v^2 = (16 + 118.523...) m^2/s^2

= 134.523... m^2/s^2, so that

v = 11.598... m/s.

Live long and prosper.

Answer 3

hi! I purely have time for the 1st one, yet extra helpful then none! #a million. H = a million.30 m V(very final) = 0 m/s (it inform you it reaches max top, so it has to come lower back to a end) D= a million.30/sin27 = 2.86m artwork = a metamorphosis in capability W = FD F = mgh mgh X d = a million/2mv(very final)2 - a million/2mv(preliminary)2 mghd = -a million/2mv(intial)2 mass cancels out ghd = -a million/2mv(intial)2 v2 = -2(-9.eighty one)(a million.30)(2.86) v(intial) = 8.fifty 5 m/s