Force and coefficient of friction problem

AI Thread Summary
To determine the acceleration of a stationary block on the floor under an applied force of 0.500mg at a 20-degree angle, a free body diagram is essential. The normal force (F_N) and frictional force (Fr) must be calculated using the coefficients of static (0.6) and kinetic friction (0.5). The challenge lies in deciding which coefficient to apply and how to eliminate the mass (m) from the equations. The equations derived include F_N + 5e-7(sin 20) = 9.8m and 5e-7(cos 20) - Fr = ma. Clarifying the use of static versus kinetic friction is crucial for solving the problem accurately.
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A initially stationary block of mass "m" is on the floor. A force of magnitude 0.500mg is then applied at upward angle 20 degrees. What is the magnitude of the acceleration of the block across the floor if coefficient of static friction = 0.6 and coefficient of kinetic friction = 0.5?

i converted 0.5 mg into 5e-7 kg. so i drew a free body diagram and made the following equations: (F_N is the normal force)
F_N + 5e-7(sin 20) = 9.8m
5e-7(cos 20) - Fr = ma.

so i solved for F_N and substituted into second equation:
since Fr = F_N (coefficient of friction) but which coefficient of friction do i use in this scenario? also even if i do find which one to use, i can't eliminate m from the equation so how do i solve for a?
 
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A force of magnitude 0.500*m*g is applied. Here m is the mass of the block.
 
rl.bhat said:
A force of magnitude 0.500*m*g is applied. Here m is the mass of the block.

thank you! i can't believe i mistook mg for milligrams...
 
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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