Simple problem in calculating kinetic energy

AI Thread Summary
To calculate the kinetic energy gained by the crate, the net force must be determined by subtracting the friction force from the applied force. The applied force is 100 N, and the friction force is 70 N, resulting in a net force of 30 N. The work done on the crate, which translates to the change in kinetic energy, is calculated by multiplying the net force by the distance traveled (10 meters), yielding 300 J. Thus, the kinetic energy gained by the crate is 300 J, not 1700 J as initially suggested. Understanding the impact of friction is crucial in accurately determining the kinetic energy.
Laurlaur790
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Homework Statement


If you push a crate horizontally with a force of 100 N across a 10 meter factory floor, and the friction between the crate the the floor is a steady 70 N, how much kinetic energy is gained by the crate?


Homework Equations


KE=1/2mv²
Work=change in KE
Work=force*distance traveled

The Attempt at a Solution



Im not sure how to get the net force. If all you have to do is add the friciton force and force from push than the solution should be:
+100-(-70)=170 N
170 N(10 meters)=1700 J

is this correct?
 
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Welcome to PF!

Hi Laurlaur790! Welcome to PF! :smile:
Laurlaur790 said:
If you push a crate horizontally with a force of 100 N across a 10 meter factory floor, and the friction between the crate the the floor is a steady 70 N, how much kinetic energy is gained by the crate?
…
If all you have to do is add the friciton force and force from push than the solution should be:
+100-(-70)=170 N
170 N(10 meters)=1700 J

is this correct?

Nooo … what effect do you think friction has on KE? :wink:
 
I think that the more friction there is, the less kinetic energy there is.
 
Laurlaur790 said:
I think that the more friction there is, the less kinetic energy there is.

That's right! :biggrin:

So what effect does that have on the numbers?
 
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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