One more Physics B problem =)?

Question

A plane 5 meter in length is inclined at an angle of 37. A block of weight 20N is placed at the top of the plane and allowed to slide down. The work done on the block by the gravitational force during the 5m slide down the plane is most nearly
A)20J
b)60J
c)80J
d)100J
e)130J

Answer 1

The change in height of the block is 5sin(37)=3 m so the change in potential energy is Fd = 20*3 = 60J which represents exactly the work done on the block by gravity.

Doug
EDIT: pleasehelpme is lost.

Answer 2

Ok, it is not so hard...so, don't panic
just draw a force diagram for the problem. Then, you have that
w=fd cos (theta)
you have your force, which is 20 N, then, you have your distance which is 5 meters, and you have your angle which is 37 from the x plate, BUT it is 127 degrees from the Y plane, where gravity is
then
w=20*5*cos (127)
the answer turns out to be -60.18 J, which is negative work done by gravity

So the answer is 60J

EDIT: oh...I thought the block was trying to go up. Then, it is 60J but not -60J. Gravity does a positive work

Answer 3

The work is done by F a force component of W (parallel to the incline)
Ew=F S =
Ew=W sin(37) S
Ew= 20 sin(37) 5=60 Joules

Answer 4

Remember Pythagorean Numbers 3, 4 and 5 and the angles 37 and 53 which add to 90 degree.
The angle opposite to smaller side viz 3 is small
and hence 37 degree.
The angle opposite to larger side viz 4 is 53 degree.
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If we remember this , in the given problem the vertical height [opposite to 37 degree is 3 and hence the work done by gravity is
20 x 3 = 60 J.
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