Force and work on a shopping cart

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The discussion focuses on calculating the work done by a shopper pushing a cart with a force of 35.0 N at a 28.0° angle over a distance of 46.0 m. The correct approach involves using the formula W = Force * displacement, specifically considering the horizontal component of the force, which is 35*cos(28). A participant points out a potential calculator error in the initial calculation, noting that simply multiplying 35 by 46 yields an incorrect result. The conversation emphasizes the importance of including the trigonometric function in the calculation to determine the accurate work done. Accurate calculations are crucial for solving physics problems effectively.
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Homework Statement


A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 28.0° downward from the horizontal. Find the work done by the shopper on the cart as he moves down an aisle 46.0 m long.

What's going on?

Homework Equations


W = Force * displacement

The Attempt at a Solution


I used the formula above to get
F=35(cos(28))46= 1610

I don't know how else to approach the problem
 

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I think your answer is correct. The vertical component of the force pushes the cart into the floor but there is no movement in that direction so it doesn't do any work. Only the 35*cos(28) component does work as you wrote.
 


According to my homework website, I'm wrong from 10%-100%. =(
 


Maybe they are wrong?

It happens.
 


Hi aaronb,

aaronb said:

Homework Statement


A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 28.0° downward from the horizontal. Find the work done by the shopper on the cart as he moves down an aisle 46.0 m long.

What's going on?

Homework Equations


W = Force * displacement


The Attempt at a Solution


I used the formula above to get
F=35(cos(28))46= 1610

I believe you are making a calculator error here. When you did the calculation, you left out the trig function, because 35 x 46=1610.
 


You're right!
 
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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