Did This Student Calculate the Physics Homework Correctly?

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The student calculated the force of sliding friction for the tree as 1097.6 N, using the coefficient of friction and the normal force derived from the tree's weight. They determined the net force acting on the tree by subtracting the frictional force from the applied force of 1200 N, resulting in a net force of 102.4 N. This led to an acceleration of 0.32 m/s² for the tree. The calculations were based on the correct application of Newton's laws and the principles of friction. Overall, the student's approach and results appear to be accurate.
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Homework Statement



A tractor exters a force of 1200N to pull a fallen tree to pull a fallen tree along the ground. If the tree has a mass of 320KG and the coefficient of sliding friction is 0.35, find
a. the force of sliding friction
b. the netforce on the tree
c. the acceleration of the tree

Note:In case of sliding friction, if the force applied to a body is greater than the force needed to overcome friciton, there is acceleration.

Two children are pulled on a sled over snow-covered ground. The sled which is initially at rest, is pulled by a rope that makes an angle of 40 degrees with the horizontal. The children have combined mass of 45 kg and the sled has a mass of 5kg. The coefficient of kinetic and static friction are U(k)=0.15 and U(s)= 0.2 . Find the frictional force exerted by the grounf on the sled and the acceleration of the children and sled, starting from rest, if the tension in the rope is a)100N b)140N


Homework Equations



Forces in the x direction is Fx: 1200 - friction = ma


The Attempt at a Solution


I think the force of sliding friction would be

1097.6

Acceleration of tree is .32m/s

force of sliding friction is the coefficient * Normal force

You can find normal force because Fy: N - Weight of tree = 0 That is the forces in the y direction.
N = weight of the tree and weight of the tree is mass times gravity.

N* .35 = 1097.6


Forces in the x direction is Fx: 1200 - friction = ma
1200-1097.6 = 320a
a = (1200-1097.6)/320
a = .32

I hope that's right >_>
 
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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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