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Physics Urgent HELP! (1 Viewer)

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Mar 15, 2008
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Hey guys,

I'm having trouble with these questions - to do with kinetic and potential energy with changing velocity

Q: A body of mass 1k is projected horizontally, with a speed of 4ms-1 from the top of a rower 40m high, and strikes the ground at C. Point B is 30m above the ground. Calculate:

a. the total energy of the body at A
b. the potential energy lost in moving from A to B
c. the kinetic energy at B
d. the kinetic energy at C

(NOTE: It has a diagram of a 40m tower with a circular motion of the mass - mass starts at point A at the top of the tower and there is a point B around the middle with a height of 30m and C which is the bottom of of the path)

Q: A body of mass 0.50 kg is projected horizontally, with a speed of 8ms-1 from the top of a cliff 45m high.

a. What is the initial kinetic energy of the body?
b. What is the kinetic energy just before striking the ground? (have no idea)


Any help would be great,
Cheers lookoutastroboy
 

QnTiria

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CHECK WITH THE ANSWERS- I MIGHT NOT BE RIGHT (forgot if energy is vector or not lulz)

a) total energy= kinetic +potential. kinetic= 0.5x m x v sq. = 0.5 x 1 x 4sq. = 8J
potential= m x g x h= 1 x 9.8x 40= 392J therefore total= 400 J

b) I DONT KNOW IF THIS APPLIES IF IT FALLS AT AN ANGLE BTW
Ep (A) - Ep (B)= m x g x 40- mx g x 30= 98J

c)no air friction? never done it this way before but- velocity down (sq)= u (sq) + 2as= 0 + 2 x 9.8 x 10 therefore v (down)= 14 m/s
if there is no air friction and other forces then newtons first law states that the mass really should still be moving 4 m/s horizontally (i think) therefore vector diagram... 4 <- + 14 (down)= 14.56 m/s W74S
Ek= 1/2 x m x v (sq)= 1/2 x 1 x 14.56 (sq)= 105.99 J

d) see prev. equation with velocity v=28 m/s and vector diagram... v = 28.28 W81S
Ek= 1/2 x 1 x 28.28 (sq)= 400 J
 

QnTiria

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Nov 8, 2009
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i think you do q 2 the same way with vector diagrams. however, please check the answers and see if my answers are actually right because i probs have done it wrong lol
 

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