Work done by gravity on block sliding down incline

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To calculate the work done by gravity on a 5.00 kg block sliding down a 30.0-degree incline, the gravitational force must be resolved into components, specifically the vertical component that acts along the incline. The work done by gravity is determined by the formula W = F * d * cos(θ), where F is the gravitational force component along the incline, d is the displacement, and θ is the angle between the force and displacement. The work done by friction, which opposes the motion, is calculated using the frictional force (dependent on the coefficient of kinetic friction) and the same displacement, while the work done by the normal force is zero since it acts perpendicular to the displacement. To find the vertical components, the sine and cosine functions of the incline angle can be used to determine the effective forces and displacements. Understanding these components is crucial for accurately calculating the work done by each force acting on the block.
TheNextOne21
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Starting from rest, a 5.00 kg block slides 2.50 m down a rough 30.0 incline. The coefficient of kinetic friction betwen the block and incline is .436

i need to find the work done by the force of gravity, work done by the friction force between block and incline and the work done by the normal force.

can anyone help me?
 
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The work done by each force is the dot product between the force and the object's displacement. For gravity, the force is at an angle to the displacement, so you'll need to take that into account. For friction, the force is against the displacement (in the opposite direction of displacement). The normal force is perpendicular to the displacement.
 


can you elaborate more on the gravity part.
 


find the vertical (y) components of the force as well as displacement
 
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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