Work and Force on a block of mass

AI Thread Summary
A 5.0 kg block is pushed 3.0 meters up a vertical wall at a constant velocity with a force applied at a 30-degree angle, while experiencing a coefficient of kinetic friction of 0.30. The work done by the applied force is calculated using W=Fdcosø, resulting in approximately 38.23 J, while the work done by gravity is determined using W=mgh. The normal force acting on the block is found using Fn=mgcosø, yielding about 42.48 N. It is suggested to draw a free body diagram to better understand the forces involved, particularly the normal force, which is independent of the block's weight. Clarification on the equations used and their correctness is sought for further understanding.
chamonix
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Homework Statement


A 5.0 kg block is pushed 3.0 meters at a constant velocity up a vertical wall by a constant florce applied at an angle of 30.0 degrees witht the horizontal. If the coefficient of kinetic friction between the block and the wall is 0.30, determine the following:
a) the work done by the force on the block
b) the work done by gravity on the block
c) the magnitude of the normal force between the block and the wall


Homework Equations


W=Fdcosø
W=mgh
Fn=mgcosø


The Attempt at a Solution


W=Fdcosø
W=mu*mgdcosø
W=.3*5*9.81*3*cos30
W=38.23J

w=mgh
w=5*9.81*3

fn=mg
fn=5*9.81*cos30
fn=42.48

I have a feeling that this is wrong. Any help would be most appreciated. Thanks.
 
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chamonix said:

Homework Statement


A 5.0 kg block is pushed 3.0 meters at a constant velocity up a vertical wall by a constant florce applied at an angle of 30.0 degrees witht the horizontal. If the coefficient of kinetic friction between the block and the wall is 0.30, determine the following:
a) the work done by the force on the block
b) the work done by gravity on the block
c) the magnitude of the normal force between the block and the wall


Homework Equations


W=Fdcosø
W=mgh
Fn=mgcosø


The Attempt at a Solution


W=Fdcosø
W=mu*mgdcosø
W=.3*5*9.81*3*cos30
W=38.23J

w=mgh
w=5*9.81*3

fn=mg
fn=5*9.81*cos30
fn=42.48

I have a feeling that this is wrong. Any help would be most appreciated. Thanks.
You should first draw a free body diagram of the block, and identify all the forces acting on the block. In particular, the Normal force is the force the wall exerts on the block, perpendicular to it. In this problem, the Normal force has nothing to do with the block's weight. Then use Newton 1 before attempting a solution to the work done by the force.
 
Thank you. I will try this.
So, would my equations for the problems be correct?
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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