A student shoves a 0.5kg block from the bottom of a friction-less incline plane. The student preforms 4 j of work and the block slides a distance s along the incline before it stops. What is the distance s.

Question

anonymous · Answer

W = F x d

anonymous · Answer

or Fdcos(theta)

anonymous · Answer

beyond that, no idea :(
If 30 degrees and g was 10 m/s/s then 4 J would lift a 0.5kg block up by 4/5 metres. 
Your block gets to 4/5 meters hight after sliding up 8/5 meters of ramp. < yahoo

anonymous · Answer

if i usw w=f*d what is the F?

anonymous · Answer

is it the force of the block?

anonymous · Answer

wait for this guy, he's sure to have it all correct

anonymous · Answer

lol ok

unklerhaukus · Answer

$$E=KE+U$$
consider the energy of the initial and final states

The initial energy is all kinetic energy
$$E_i=KE$$

The final energy is all gravitational potential energy 
$$E_f=U$$

unklerhaukus · Answer

all the work has gone to lifting the block

anonymous · Answer

yup so if all the kinetic energy at the botom is 4j then at the top the potential energy should then be 4j as the block has come to a stop. Do i use only the Ek = 0.5mv^2 and the Ep=mgh equations? How can i when i dont know the h or the v?

unklerhaukus · Answer

the block now has 4 j more potential gravitational energy

unklerhaukus · Answer

actually we dont need to find KE because we know the amount of energy that was used

anonymous · Answer

ok?

unklerhaukus · Answer

$$U=mgh$$

solve for the hight reached , h

anonymous · Answer

0.816 m

anonymous · Answer

now i use sin rule to find the angular displacement. THANK YOU!

unklerhaukus · Answer

then we have to draw a picture of a incline 
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