What is the acceleration of a block on a frictionless surface?

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A block on a frictionless surface experiences a force of 20 N at a 30-degree angle, and its mass is 5 kg. The horizontal component of the force is calculated as 20 cos(30), resulting in approximately 17 N. The acceleration is derived from this horizontal force divided by the mass, leading to a calculation of 3.46 m/s², which does not match the provided answer choices. Participants in the discussion confirm that the method is correct, but there may be an error in the answer options. The conversation highlights the importance of double-checking calculations in physics problems.
dirtybirdhouse183
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Homework Statement



A block is on a frictionless horizontal surface. A force of 20 N is applied to the block at an angle of 30 degrees to the horizontal. If the mass of the block is 5 kg, what is the acceleration of the block?

The choices for acceleration are: a) 5.3 b) 6.2 c) 7.5 d) 4.7 e)3.2

Homework Equations



Do I need to compute the vertical forces acting on the block?

The Attempt at a Solution



So I am guessing that the horizontal component of the force is the force that causes the block to accelerate, or move. So then the horizontal force would be 20cos30 which is 17N. Then I divided that by the mass to get the acceleration but it's wrong.
 
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I did what you did:
F_x = \cos 30 \cdot 20 = m a_x
F_y = \sin 30 \cdot 20 - n = 0
so a_x = \frac {\cos 30 \cdot 20}{5} = .62
 
dirtybirdhouse183 said:
So I am guessing that the horizontal component of the force is the force that causes the block to accelerate, or move. So then the horizontal force would be 20cos30 which is 17N. Then I divided that by the mass to get the acceleration but it's wrong.
Your method is correct; the answer is not among those given (although one of them is pretty close).

lizzyb said:
so a_x = \frac {\cos 30 \cdot 20}{5} = .62
Check your arithmetic.
 
I tried

Yes, my acceleration was 3.46, but there acceleration is 3.2. Hopefully, the answer choices aren't incorrect lol. Thanks for the replies though, I really apperciate them.
 
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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