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Mechanics 4 problem

The question is: A block of mass 60 kg is pulled up a hill in the line of greatest slope by a force of magnitude 50 N acting at an angle X above the hill. The block passes through points A and B with speeds 8.5 m s^−1 and 3.5 m s^−1 respectively. The distance AB is 250 m and B is 17.5 m above the level of A. The resistance to motion of the block is 6 N. Find the value of X.

OK. So I used the constant acceleration formula v^2=u^2+2as to find the acceleration of the block as it moves up the hill, which I calculated as -0.12 ms^-2.

Then I resolved the driving force of 50 N as 50cosX, parallel to the hill, and 50sinX. The resistive force of 6 N is parallel to the hill as well, and I need to find out the component of the weight of the block (600 N) that is also parallel to the slope. Now this where I'm stuck.

What is the angle that I should use for the the weight component parallel to the slope? I tried using trigonometry to calculate that only to realize I need an exact value for questions like these. I attempted to use 600sin(tan^-1(17.5/250)) for an exact answer for X but I got something like 35.6 instead of 35.3, the actual answer. Where did I go wrong?

This is how I went about it:
50cosX - 6 - (?)600(sin(tan^-1(17.5/250)))(?) = 60(-0.12)

(edited 7 years ago)

Reply 1

JN17

I don't do M4 however points A and B are points on a slope, meaning AB is the hypotenuse of a triangle so should you not be using sin^-1(17.5/250) to find the angle? Could just be a difference in diagrams but doing this gives you cos^-1(0.816) as your final answer which is 35.3 to 3sf.

Reply 2

Heman4002

Bro don't use tan^-1(17.5/250)Instead We know the height 17.5 and the AB 250 m So a triangle forms now find {[sin]} Now while resolving forces it will be 50cosX -(6 600sintheta)=60×(-0.12)Therefore now. 50cosX -(6 (600×17.5/250))=-7.250cosX -(6 42)=-7.250coX =48-7.2CosX= 40.8/50X=cos^-1(40.8/50)X=35.3