How Much Force to Accelerate a Box with Known Friction?

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To determine the horizontal force required to accelerate a 4.0 kg box with a coefficient of friction of 0.60 at 2.0 m/s², first calculate the friction force using the formula: friction force = coefficient of friction * normal force. The normal force equals the weight of the box, which is 4.0 kg multiplied by gravity (9.81 m/s²), resulting in a normal force of approximately 39.24 N. The total force needed can be found by applying Newton's second law (F = ma), which gives a force of 8 N for acceleration. Adding the friction force to this total yields the necessary horizontal force to overcome friction and achieve the desired acceleration.
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I need help on this question, The coefficient of friction between a 4.0 kg box and the floor is 0.60. How large of a horizontal force is necessary to give the box an acceleration of 2.0m/s²?
 
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that sounds suspiciously like a homework problem
 
Friction force = (coefficient of friction) * (normal force)
F = ma

all you need
 
thowonasweaterandgo said:
that sounds suspiciously like a homework problem

Yeah ! Movin' it...:smile:
 
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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