Force Applied at an Angle to an Object on Frictionless Surface

AI Thread Summary
The problem involves a block sliding on a frictionless surface with a force of 20 N applied at a 30° angle. The relevant equation is F=ma, where the force components must be considered. The user initially calculated the acceleration as 3.46 m/s² by only accounting for the horizontal component of the force. However, they overlooked the total force applied, leading to the correct acceleration being 7.5 m/s². The discussion highlights the importance of considering all forces when solving physics problems.
jpsmith394
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Homework Statement


The horizontal surface on which the block slides is frictionless. If F = 20 N and M = 5.0 kg, what is the magnitude of the resulting acceleration of the block? The force is applied 30° to the horizontal.


Homework Equations


F=ma


The Attempt at a Solution


I began by ignoring the weight and normal force in the vertical axis (y-axis) because the surface is frictionless. Then my next step is to separate the force into compenets with F_{x}=Fcos\theta

Therefore:
ƩF_{x}=Fcos\theta = ma_{x}

So:
a_{x}=\frac{Fcos\theta}{m}

And using the given numbers I got a solution of 3.46\frac{m}{s^2}
But the online system I am using shows the correct answer as 7.5\frac{m}{s^2}
 

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hi jpsmith394! welcome to pf! :smile:

erm :redface: … you forgot the other F, in front! :wink:
 
tiny-tim said:
hi jpsmith394! welcome to pf! :smile:

erm :redface: … you forgot the other F, in front! :wink:

Oh gosh I see now, I was just stuck thinking it was just a single force

Thanks,
James
 
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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