How Is Work Calculated When Lifting and Carrying a Cement Block?

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To calculate the work done when lifting and carrying a cement block, first, consider the vertical lift of 5.5 kg over 1.2 m, which involves the force of gravity. The work done by the person in lifting the block is calculated using the formula W = F × d, resulting in 64.2 joules. While carrying the block horizontally for 7.3 m, the work done against gravity is zero since there is no vertical displacement. The force of gravity does negative work during the lift, also totaling -64.2 joules. Overall, the work done by the person is positive during the lift and zero during the horizontal carry, while gravity's work is negative only during the lift.
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A person lifts a 5.5 kg cement block a vertical distance 1.2 m and then carries the block horizontally a distance 7.3 m. Determine the work done by the person and by the force of gravity in this process. How much work is done by the person and the force of gravity?

Would you break this up into components?
 
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Its not necessary if you recall the definition of work to involve a dot product.
 
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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