# M2 Work Energy PrincipleWatch

#1
let's say a block of 20kg is moving up slope at angle of 30 degrees.

starts at 8ms-1 and ends up at 2ms-1 after travelling x metres up the slope. The friction is constant 6g N. Find x.

loss in KE = gain in GPE + work done against friction

0.5*20*(8^2 - 2^2) = 20g*xsin(30) + 6gx
600/(20gsin(30) + 6g)=x
x = 3.82m

My question is why isn't the work done to overcome weight component of block parallel to the slope.

so shouldn't it be:
loss in KE = gain in GPE + work done against friction + work done against weight component = 20g*sin(30)
0
2 years ago
#2
(Original post by Bealzibub)
let's say a block of 20kg is moving up slope at angle of 30 degrees.

starts at 8ms-1 and ends up at 2ms-1 after travelling x metres up the slope. The friction is constant 6g N. Find x.

loss in KE = gain in GPE + work done against friction

0.5*20*(8^2 - 2^2) = 20g*xsin(30) + 6gx
600/(20gsin(30) + 6g)=x
x = 3.82m

My question is why isn't the work done to overcome weight component of block parallel to the slope.

so shouldn't it be:
loss in KE = gain in GPE + work done against friction + work done against weight component = 20g*sin(30)
"Work done against weight component" - that's precisely what GPE is.

If you do gain in GPE + work done against weight component, you're adding GPE to itself twice.
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