A block is placed on a inclined plane.
The acceleration of gravity is 9.8 m/s2 The acceleration of gravity is 9.8 m/s2.
A 25 N force is required to push the block
up the incline with constant velocity.
What is the weight of the block? Answer in
units of N.
2007-10-17 10:47:49 · 3 answers · asked by PhysicsDumbass 1 in Science & Mathematics ➔ Physics
It'd help (rather a lot, in fact) if one knew the angle that the inclined plane made with the horizontal ☺
Doug
2007-10-17 10:53:48 · answer #1 · answered by doug_donaghue 7 · 1⤊ 0⤋
o.k. this one is doable -
say your plane is inclined theta degrees above horizontal and the mass of the block is m so the weight is 9.8*m Newtons
if muk is the coefficient of sliding friction (also known as kinetic friction) at the constant velocity that we are talking about then the component of the weight down the plane is 9.8*m*sin(theta) and the friction force down the plane is 9.8*cos(theta)*muk so 25 = 9.8(m*sin(theta)+muk*cos(theta))
(25/9.8-muk*cos(theta))/sin(theta) = m so..
25/sin(theta) -9.8*muk /tan(theta) = weight you are looking for
(I hope, if I haven't messed anything up :( )
2007-10-17 18:34:40 · answer #2 · answered by Anonymous · 0⤊ 0⤋
there is something missing on the question ,
if not , in your case it will be 25 N
2007-10-17 18:13:52 · answer #3 · answered by MiMi 1 · 0⤊ 0⤋